Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.2

Question 1.
Evaluate:
(i) 8!
Solution:
8!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320

(ii) 10!
Solution:
10!
= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 3628800

(iii) 10! – 6!
Solution:
10! – 6!
= 10 × 9 × 8 × 7 × 6! – 6!
= 6! (10 × 9 × 8 × 7 – 1)
= 6! (5040 – 1)
= 6 × 5 × 4 × 3 × 2 × 1 × 5039
= 3628080

(iv) (10 – 6)!
Solution:
(10 – 6)!
= 4!
= 4 × 3 × 2 × 1
= 24

Question 2.
Compute:
(i) \(\frac{12 !}{6 !}\)
Solution:
\(\frac{12 !}{6 !}=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\)
= 12 × 11 × 10 × 9 × 8 × 7
= 665280

(ii) \(\left(\frac{12}{6}\right) !\)
Solution:
\(\left(\frac{12}{6}\right) !\)
= 2!
= 2 × 1
= 2

(iii) (3 × 2)!
Solution:
(3 × 2)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

(iv) 3! × 2!
Solution:
3! × 2!
= 3 × 2 × 1 × 2 × 1
= 12

(v) \(\frac{9 !}{3 ! 6 !}\)
Solution:
\(\frac{9 !}{3 ! 6 !}=\frac{9 \times 8 \times 7 \times 6 !}{(3 \times 2 \times 1) \times 6 !}=84\)

(vi) \(\frac{6 !-4 !}{4 !}\)
Solution:
\(\frac{6 !-4 !}{4 !}=\frac{6 \times 5 \times 4 !-4 !}{4 !}=\frac{4 !(6 \times 5-1)}{4 !}=29\)

(vii) \(\frac{8 !}{6 !-4 !}\)
Solution:
\(\frac{8 !}{6 !-4 !}=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{6 \times 5 \times 4 !-4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !(6 \times 5-1)}\)
= \(\frac{1680}{29}\)
= 57.93

(viii) \(\frac{8 !}{(6-4) !}\)
Solution:
\(\frac{8 !}{(6-4) !}=\frac{8 !}{2 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 20160

Question 3.
Write in terms of factorials
(i) 5 × 6 × 7 × 8 × 9 × 10
Solution:
5 × 6 × 7 × 8 × 9 × 10 = 10 × 9 × 8 × 7 × 6 × 5
Multiplying and dividing by 4!, we get
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 !}\)
= \(\frac{10 !}{4 !}\)

(ii) 3 × 6 × 9 × 12 × 15
Solution:
3 × 6 × 9 × 12 × 15
= 3 × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5)
= (3⁵) (5 × 4 × 3 × 2 × 1)
= 3⁵ (5!)

(iii) 6 × 7 × 8 × 9
Solution:
6 × 7 × 8 × 9 = 9 × 8 × 7 × 6
Multiplying and dividing by 5!, we get
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 !}\)
= \(\frac{9 !}{5 !}\)

(iv) 5 × 10 × 15 × 20
Solution:
5 × 10 × 15 × 20
= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4)
= (5⁴) (4 × 3 × 2 × 1)
= (5⁴) (4!)

Question 4.
Evaluate: \(\frac{n !}{r !(n-r) !}\) for
(i) n = 8, r = 6
(ii) n = 12, r = 12
(iii) n = 15, r = 10
(iv) n = 15, r = 8
Solution:

Question 5.
Find n, if
(i) \(\frac{n}{8 !}=\frac{3}{6 !}+\frac{1 !}{4 !}\)
Solution:

(ii) \(\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}\)
Solution:

(iii) \(\frac{1 !}{n !}=\frac{1 !}{4 !}-\frac{4}{5 !}\)
Solution:

(iv) (n + 1)! = 42 × (n -1)!
Solution:
(n + 1)! = 42(n – 1)!
∴ (n + 1) n (n – 1)! = 42(n – 1)!
∴ n² + n = 42
∴ n² + n – 42 = 0
∴ (n + 7)(n – 6) = 0
∴ n = -7 or n = 6
But n ≠ -7 as n ∈ N
∴ n = 6

(v) (n + 3)! = 110 × (n + 1)!
Solution:
(n + 3)! = (110) (n + 1)!
∴ (n + 3)(n + 2)(n + 1)! = 110(n + 1)!
∴ (n + 3) (n + 2) = (11) (10)
Comparing on both sides, we get
n + 3 = 11
∴ n = 8

Question 6.
Find n, if:
(i) \(\frac{(17-n) !}{(14-n) !}=5 !\)
Solution:

∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4
Comparing on both sides, we get
17 – n = 6
∴ n = 11

(ii) \(\frac{(15-n) !}{(13-n) !}=12\)
Solution:
\(\frac{(15-n) !}{(13-n) !}=12\)
∴ \(\frac{(15-n)(14-n)(13-n) !}{(13-n) !}=12\)
∴ (15 – n) (14 – n) = 4 × 3
Comparing on both sides, we get
∴ 15 – n = 4
∴ n = 11

(iii) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3\)
Solution:

∴ 12 = (n – 3)(n – 4)
(n – 3)(n – 4) = 4 × 3
Comparing on both sides, we get
n – 3 = 4
∴ n = 7

(iv) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-7) !}=1: 6\)
Solution:

∴ 120 = (n – 3)(n – 4) (n – 5)(n – 6)
∴ (n – 3)(n – 4) (n – 5)(n – 6) = 5 × 4 × 3 × 2
Comparing on both sides, we get
n – 3 = 5
∴ n = 8

(v) \(\frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}=24: 1\)
Solution:


(2n – 1)(2n – 3)(2n – 5) = \(\frac{24 \times 7 \times 6 \times 5}{16}\)
∴ (2n – 1)(2n – 3)(2n – 5) = 9 × 7 × 5
Comparing on both sides. We get
∴ 2n – 1 = 9
∴ n = 5

Question 7.
Show that \(\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !}=\frac{(n+1) !}{r !(n-r+1) !}\)
Solution:

Question 8.
Show that \(\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{10 !}{4 ! 6 !}\)
Solution:

Question 9.
Show that \(\frac{(2 n) !}{n !}\) = 2n (2n – 1)(2n – 3)…5.3.1
Solution:

Question 10.
Simplify
(i) \(\frac{(2 n+2) !}{(2 n) !}\)
Solution:

(ii) \(\frac{(n+3) !}{\left(n^{2}-4\right)(n+1) !}\)
Solution:

(iii) \(\frac{1}{n !}-\frac{1}{(n-1) !}-\frac{1}{(n-2) !}\)
Solution:

(iv) n[n! + (n – 1)!] + n²(n – 1)! + (n + 1)!
Solution:

(v) \(\frac{n+2}{n !}-\frac{3 n+1}{(n+1) !}\)
Solution:

(vi) \(\frac{1}{(n-1) !}+\frac{1-n}{(n+1) !}\)
Solution:

(vii) \(\frac{1}{n !}-\frac{3}{(n+1) !}-\frac{n^{2}-4}{(n+2) !}\)
Solution:

(viii) \(\frac{n^{2}-9}{(n+3) !}+\frac{6}{(n+2) !}-\frac{1}{(n+1) !}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.1

Question 1.
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can the monitor be selected if the monitor must be a girl?
Solution:
There are 30 boys and 20 girls.
A teacher can select any boy as a class monitor from 30 boys in 30 different ways and he can select any girl as a class monitor from 20 girls in 20 different ways.
∴ by the fundamental principle of addition, the total number of ways a teacher can select a class monitor = 30 + 20 = 50
Hence, there are 50 different ways to select a class monitor.

Question 2.
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Solution:
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication, total no. of ways a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.

Question 3.
How many two-letter words can be formed using letters from the word SPACE, when repetition of letters (i) is allowed (ii) is not allowed
Solution:
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed 1st letter can be selected in 5 ways.
2nd letter can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, the total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed 1st letter can be selected in 5 ways.
2nd letter can be selected in 4 ways.
∴ By using the fundamental principle of multiplication, the total number of 2-letter words = 5 × 4 = 20

Question 4.
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits (i) are allowed (ii) are not allowed?
Solution:
A three-digit number is to be formed from the digits 0, 1, 3, 5, 6.
(i) When repetition of digits is allowed,
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5,6.
∴ 100’s place digit can be selected in 4 ways.
10’s and unit’s place digit can be zero and digits can be repeated.
∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, the total number of three-digit numbers = 4 × 5 × 5 = 100

(ii) When repetition of digits is not allowed,
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6.
∴ 100’s place digit can be selected in 4 ways.
10’s and unit’s place digit can be zero and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways.
∴ By using the fundamental principle of multiplication, the total number of two-digit numbers = 4 × 4 × 3 = 48

Question 5.
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Solution:
A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
∴ Unit’s place digit can be selected in 5 ways.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 5 ways.
∴ By using fundamental principle of multiplication, total number of 3-digit numbers = 5 × 5 × 5 = 125

Question 6.
A letter lock contains 3 rings and each ring contains 5 letters. Determine the maximum number of false trails that can be made before the lock is opened.
Solution:
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, a total number of trials that can be made = 5 × 5 × 5 = 125.
Out of these 124 wrong attempts are made and in the 125th attempt, the lock gets opened.
∴ A maximum number of false trials = 124.

Question 7.
In a test, 5 questions are of the form ‘state, true or false. No student has got all answers correct. Also, the answer of every student is different. Find the number of students who appeared for the test.
Solution:
Every question can be answered in 2 ways. (True or False)
∴ By using the fundamental principle of multiplication, the total number of set of answers possible = 2 × 2 × 2 × 2 × 2 = 32.
Since One of them is the case where all questions are answered correctly,
The number of wrong answers = 32 – 1 = 31.
Since no student has answered all the questions correctly, the number of students who appeared for the test are 31.

Question 8.
How many numbers between 100 and 1000 have 4 in the unit’s place?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where the unit place digit is 4.
Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For a 3-digit number, 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using the fundamental principle of multiplication,
total numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90.

Question 9.
How many numbers between 100 and 1000 have the digit 7 exactly once?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
When 7 is in unit’s place:
Unit’s place digit is 7.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 9 ways.
100’s place digit can be selected in 8 ways.
Total numbers which have 7 in unit’s place = 1 × 9 × 8 = 72.

When 7 is in 10’s place:
Unit’s place digit can be selected in 9 ways.
10’s place digit is 7.
∴ it can be selected in 1 way only.
100’s place digit can be selected in 8 ways.
∴ A total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72.

When 7 is in 100’s place:
Unit’s place digit can be selected in 9 ways.
10’s place digit can be selected in 9 ways.
100’s place digit is 7.
∴ it can be selected in 1 way only.
∴ Total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81.
∴ Total numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225

Question 10.
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Solution:
Between any set of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed,
1000’s place digit can be selected in 4 ways,
100’s place digit can be selected in 3 ways,
10’s place digit can be selected in 2 ways,
Unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed with the digits 2, 3, 4, 7 = Total number of four-digit numbers possible from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24

Question 11.
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution:
Case I: Number of three-digit numbers formed from 2, 3, 4, 5, 6, greater than 400.
100’s place can be filled by any one of the numbers 4, 5, 6.
100’s place digit can be selected in 3 ways.
Since repetition is not allowed, 10’s place can be filled by any one of the remaining four numbers.
∴ 10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ Total number of three-digit numbers formed = 3 × 4 × 3 = 36

Case II: Number of four-digit numbers formed from 2, 3, 4, 5, 6.
Since repetition of digits is not allowed,
1000’s place digit can be selected in 5 ways.
100’s place digit can be selected in 4 ways.
10’s place digit can be selected in 3 ways.
Unit’s place digit can be selected in 2 ways.
∴ Total number of four-digit numbers formed = 5 × 4 × 3 × 2 = 120

Case III: Number of five-digit numbers formed from 2, 3, 4, 5, 6.
Similarly, since repetition of digits is not allowed,
Total number of five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120.
∴ Total number of numbers that exceed 400 = 36 + 120 + 120 = 276

Question 12.
How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Solution:
Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8.
10’s place digit is 1.
∴ it can be selected in 1 way only.
Unit’s place can be filled by any one of the numbers 5, 7, 8.
∴ Unit’s place digit can be selected in 3 ways.
∴ Total number of such numbers = 1 × 3 = 3.

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8.
10’s place can be filled by any one of the numbers 2, 5, 7, 8.
∴ 10’s place digit can be selected in 4 ways.
Since repetition is allowed, the unit’s place can be filled by one of the remaining 6 digits.
∴ Unit’s place digit can be selected in 6 ways.
∴ Total number of such numbers = 4 × 6 = 24.

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8.
Similarly, since repetition of digits is allowed, the total number of such numbers = 5 × 6 × 6 = 180.
All cases are mutually exclusive.
∴ Total number of required numbers = 3 + 24 + 180 = 207

Question 13.
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Solution:
A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ A number of ways to choose gates = 3.
The number of ways to choose a staircase = 4.
By using the fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12

Question 14.
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 5 if digits are not repeated?
Solution:
Here, repetition of digits is not allowed.
For a number to be divisible by 5,
unit’s place digit should be 0 or 5.
Case I: when unit’s place is 0
Unit’s place digit can be selected in 1 way.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 4 ways.
1000’s place digit can be selected in 3 ways.
10000’s place digit can be selected in 2 ways.
∴ Total number of numbers = 1 × 5 × 4 × 3 × 2 = 120.

Case II: when the unit’s place is 5
Unit’s place digit can be selected in 1 way.
10000’s place should be a non-zero number.
∴ It can be selected in 4 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
∴ Total number of numbers = 1 × 4 × 4 × 3 × 2 = 96
∴ Total number of required numbers = 120 + 96 = 216

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

(I) Select the correct answer from the given alternative:

Question 1.
The common ratio for the G.P. 0.12, 0.24, 0.48, is
(A) 0.12
(B) 0.2
(C) 0.02
(D) 2
Answer:
(D) 2

Question 2.
The tenth term of the geometric sequence is \(\frac{1}{4}, \frac{-1}{2}, 1,-2, \ldots\) is
(A) 1024
(B) \(\frac{1}{1024}\)
(C) -128
(D) \(\frac{-1}{128}\)
Answer:
(C) -128
Hint:

Question 3.
If for a G.P. \(\frac{t_{6}}{t_{3}}=\frac{1458}{54}\) then r = ?
(A) 3
(B) 2
(C) 1
(D) -1
Answer:
(A) 3
Hint:

Question 4.
Which term of the geometric progression 1, 2, 4, 8, ….. is 2048.
(A) 10th
(B) 11th
(C) 12th
(D) 13th
Answer:
(C) 12th
Hint:
Here, a = 1, r = 2
nth term of geometric progression = ar^n-1
∴ ar^n-1 = 2048
2^n-1 = 2¹¹
n – 1 = 11
∴ n = 12

Question 5.
If the common ratio of the G.P. is 5, the 5th term is 1875, the first term is
(A) 3
(B) 5
(C) 15
(D) -5
Answer:
(A) 3

Question 6.
The sum of 3 terms of a G.P. is \(\frac{21}{4}\) and their product is 1, then the common ratio is
(A) 1
(B) 2
(C) 4
(D) 8
Answer:
(C) 4
Hint:
Let three terms be \(\frac{a}{r}\), a, ar
According to the given conditions,
\(\frac{a}{r}\) + a + ar = \(\frac{21}{4}\) …..(i)
and \(\frac{a}{r}\) × a × ar = 1,
i.e., a³ = 1
∴ a = 1
∴ from equation (i), we get
\(\frac{1}{r}\) + 1 + r = \(\frac{21}{4}\)
By solving this, we get r = 4.

Question 7.
Sum to infinity of a G.P. 5, \(-\frac{5}{2}, \frac{5}{4},-\frac{5}{8}, \frac{5}{16}, \ldots\) is
(A) 5
(B) \(-\frac{1}{2}\)
(C) \(\frac{10}{3}\)
(D) \(\frac{3}{10}\)
Answer:
(C) \(\frac{10}{3}\)
Hint:
Here, a = 5, r = \(\frac{-1}{2}\), |r| < 1
∴ Sum to the infinity = \(\frac{a}{1-r}=\frac{5}{1+\frac{1}{2}}=\frac{10}{3}\)

Question 8.
The tenth term of H.P. \(\frac{2}{9}, \frac{1}{7}, \frac{2}{19}, \frac{1}{12}, \ldots\) is
(A) \(\frac{1}{27}\)
(B) \(\frac{9}{2}\)
(C) \(\frac{5}{2}\)
(D) 27
Answer:
(A) \(\frac{1}{27}\)
Hint:

Question 9.
Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively, (a, b > 0)
(A) A = \(\frac{a+b}{2}\)
(B) G = \(\sqrt{a b}\)
(C) H = \(\frac{2 a b}{a+b}\)
(D) A = GH
Answer:
(D) A = GH

Question 10.
The G.M. of two numbers exceeds their H.M. by \(\frac{6}{5}\), the A.M. exceeds G.M. by \(\frac{3}{2}\) the two numbers are
(A) 6, \(\frac{15}{2}\)
(B) 15, 25
(C) 3, 12
(D) \(\frac{6}{5}\), \(\frac{3}{2}\)
Answer:
(C) 3, 12
Hint:

(II) Answer the following:

Question 1.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Solution:

Question 2.
Find the sum of the first 5 terms of the G.P. whose first term is 1 and the common ratio is \(\frac{2}{3}\).
Solution:

Question 3.
For a G.P. a = \(\frac{4}{3}\) and t₇ = \(\frac{243}{1024}\), find the value of r.
Solution:

Question 4.
For a sequence, if \(t_{n}=\frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Solution:
The sequence (t_n) is a G.P.
if \(\frac{\mathrm{t}_{\mathrm{n}+1}}{\mathrm{t}_{\mathrm{n}}}\) = constant for all n ∈ N.

Question 5.
Find three numbers in G.P. such that their sum is 35 and their product is 1000.
Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar.
According to the given conditions,
\(\frac{a}{r}\) + a + ar = 35
a(\(\frac{1}{r}\) + 1 + r) = 35 …..(i)
Also, (\(\frac{a}{r}\))(a)(ar) = 1000
a³ = 1000
∴ a = 10
Substituting the value of a in (i), we get

Hence, the three numbers in G.P. are 20, 10, 5, or 5, 10, 20.

Question 6.
Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Solution:
Let the five numbers in G.P. be \(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\).
According to the given condition,

Question 7.
For a sequence, S_n = 4(7ⁿ – 1), verify that the sequence is a G.P.
Solution:

∴ The given sequence is a G.P.

Question 8.
Find 2 + 22 + 222 + 2222 + … upto n terms.
Solution:
S_n = 2 + 22 + 222 +… upto n terms
= 2(1 + 11 + 111 + ….. upto n terms)
= \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) + …… upto n terms]
= \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)]
Since 10, 100, 1000, ….. n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10,

Question 9.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…
Solution:
0.6, 0.66, 0.666, 0.6666, …
∴ t₁ = 0.6
t₂ = 0.66 = 0.6 + 0.06
t₃ = 0.666 = 0.6 + 0.06 + 0.006
Hence, in general
t_n = 0.6 + 0.06 + 0.006 + …..upto n terms.
The terms are in G.P. with
a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1
∴ t_n = the sum of first n terms of the G.P.

Question 10.
Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\)
Solution:

Question 11.
Find \(\sum_{r=1}^{n} r(r-3)(r-2)\)
Solution:

Question 12.
Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)
Solution:
We know that

Question 13.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots r^{3}}{(r+1)^{2}}\)
Solution:

Question 14.
Find 2 × 6 + 4 × 9 + 6 × 12 + ….. upto n terms.
Solution:
2, 4, 6, ….. are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
6, 9, 12, ….. are in A.P.
∴ rth term = 6 + (r – 1)(3) = (3r + 3)
∴ 2 × 6 + 4 × 9 + 6 × 12 + ….. to n terms

= n(n + 1) [2n + 1 + 3]
= 2n(n + 1)(n + 2)

Question 15.
Find 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + …… upto n terms.
Solution:
2, 4, 6,… are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
5, 7, 9, … are in A.P.
∴ rth term = 5 + (r – 1) (2) = (2r + 3)
8, 10, 12, … are in A.P.
∴ rth term = 8 + (r – 1) (2) = (2r + 6)
2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + ….. to n terms

= 2n (n + 1) [n(n + 1) + 3(2n + 1) + 9]
= 2n (n + 1)(n² + 7n + 12)
= 2n (n + 1) (n + 3) (n + 4)

Question 16.
Find \(\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{2}+\frac{1^{2}+2^{2}+3^{2}}{3}+\ldots\) upto n terms.
Solution:

Question 17.
Find 12² + 13² + 14² + 15² + ….. 20²
Solution:

Question 18.
If \(\frac{1+2+3+4+5+\ldots \text { upto } \mathrm{n} \text { terms }}{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{3}{22}\), Find the value of n.
Solution:

Question 19.
Find (50² – 49²) + (48² – 47²) + (46² – 45²) +… + (2² – 1²).
Solution:

Question 20.
If \(\frac{1 \times 3+2 \times 5+3 \times 7+\ldots \text { upto } \mathrm{n} \text { terms }}{1^{3}+2^{3}+3^{3}+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{5}{9}\), find the value of n.
Solution:

Question 21.
For a G.P. if t₂ = 7, t₄ = 1575, find a.
Solution:

Question 22.
If for a G.P. t₃ = \(\frac{1}{3}\), t₆ = \(\frac{1}{81}\) find r.
Solution:

Question 23.
Find \(\sum_{r=1}^{n}\left(\frac{2}{3}\right)^{r}\).
Solution:

Question 24.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
Solution:
Since k – 1, k, k + 2 are consecutive terms of a G.P.,
\(\frac{k}{k-1}=\frac{k+2}{k}\)
k² = k² + k – 2
k – 2 = 0
∴ k = 2

Question 25.
If for a G.P. first term is (27)² and the seventh term is (8)², find S₈.
Solution:

Question 26.
If pth, qth and rth terms of a G.P. are x, y, z respectively. Find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Solution:
Let a be the first term and R be the common ratio of the G.P.

Question 27.
Which 2 terms are inserted between 5 and 40 so that the resulting sequence is G.P.
Solution:
Let the required numbers be G₁ and G₂.

∴ For the resulting sequence to be in G.P. we need to insert numbers 10 and 20.

Question 28.
If p, q, r are in G.P. and \(\mathrm{p}^{1 / \mathrm{x}}=\mathrm{q}^{1 / \mathrm{y}}=\mathrm{r}^{1 / \mathrm{z}}\), verify whether x, y, z are in A.P. or G.P. or neither.
Solution:

Question 29.
If a, b, c are in G.P. and ax² + 2bx + c = 0 and px² + 2qx + r = 0 have common roots, then verify that pb² – 2qba + ra² = 0.
Solution:
a, b, c are in G.P.
∴ b² = ac
ax² + 2bx + c = 0 becomes

Question 30.
If p, q, r, s are in G.P., show that (p² + q² + r²)(q² + r² + s²) = (pq + qr + rs)².
Solution:
p, q, r, s are in G.P.

Question 31.
If p, q, r, s are in G.P., show that (pⁿ + qⁿ), (qⁿ + rⁿ), (rⁿ + sⁿ) are also in G.P.
Solution:
p, q, r, s are in G.P.
Let the common ratio be R
∴ let p = \(\frac{a}{R^{3}}\), q = \(\frac{a}{R}\), r = aR and s = aR³
To show that (pⁿ + qⁿ), (qⁿ + rⁿ), (rⁿ + sⁿ) are in G.P,
i.e., we have to show

Question 32.
Find the coefficient x⁶ in the expression of e^2x using series expansion.
Solution:

Question 33.
Find the sum of infinite terms of \(1+\frac{4}{5}+\frac{7}{25}+\frac{10}{125}+\frac{13}{625}+\ldots\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.6

Question 1.
Find the sum \(\sum_{r=1}^{n}(r+1)(2 r-1)\).
Solution:

Question 2.
Find \(\sum_{r=1}^{n}\left(3 r^{2}-2 r+1\right)\)
Solution:

Question 3.
Find \(\sum_{r=1}^{n}\left(\frac{1+2+3 \ldots .+r}{r}\right)\)
Solution:

= \(\frac{n}{4}\) [(n + 1) + 2]
= \(\frac{n}{4}\) (n + 3)

Question 4.
Find \(\sum_{r=1}^{n}\left(\frac{1^{3}+2^{3}+\ldots . .+r^{3}}{r(r+1)}\right)\)
Solution:

Question 5.
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms.
Solution:
5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms
Now, 5, 9, 13, … are in A.P. with
rth term = 5 + (r – 1) (4) = 4r + 1
7, 11, 15, ….. are in A.P. with
rth term = 7 + (r – 1) (4) = 4r + 3
∴ 5 × 7 + 9 × 11 + 13 × 15 + …… upto n terms

Question 6.
Find the sum 2² + 4² + 6² + 8² + ….. upto n terms.
Solution:

Question 7.
Find (70² – 69²) + (68² – 67²) + (66² – 65²) + …… + (2² – 1²)
Solution:
Let S = (70² – 69²) + (68² – 67²) + …… + (2² – 1²)
∴ S = (2² – 1²) + (4² – 3²) + ….. + (70² – 69²)
Here, 2, 4, 6,…, 70 are in A.P. with rth term = 2r
and 1, 3, 5, …,69 are in A.P. with rth term = 2r – 1

Question 8.
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution:
Let S = 1 × 3 × 5 + 3 × 5 × 7 + ….. upto n terms
Here, 1, 3, 5, 7 … are in A.P. with rth term = 2r – 1,
3, 5, 7, 9,… are in A.P. with rth term = 2r + 1,
5, 7, 9, 11,… are in A.P. with rth term = 2r + 3

Question 9.
If \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } n \text { terms }}{1+2+3+4+\ldots \text { upto } n \text { terms }}\) = \(\frac{100}{3}\), find n.
Solution:

Question 10.
If S₁, S₂ and S₃ are the sums of first n natural numbers, their squares and their cubes respectively, then show that 9\(\mathrm{S}_{2}{ }^{2}\) = S₃(1 + 8S₁).
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.5

Question 1.
Find Sn of the following arithmetico-geometric sequences.
(i) 2, 4x, 6x², 8x³, 10x⁴, ……
Solution:
2, 4x, 6x², 8x³, 10x⁴, ……
Here, 2, 4, 6, 8, 10,… are in A.P.
∴ a = 2, d = 2
∴ nth term = a + (n – 1)d
= 2 + (n – 1)(2)
= 2n

(ii) 1, 4x, 7x², 10x³, 13x⁴, ……
Solution:
1, 4x, 7x², 10x³, 13x⁴, ……
Here, 1, 4, 7, 10, 13,… are in A.P.
a = 1, d = 3
∴ nth term = a + (n – 1)d
= 1 + (n – 1)(3)
= 3n – 2
Also, 1, x, x², x³,… are in G.P.
∴ a = 1, r = x,
nth term = ar^n-1 = x^n-1
nth term of arithmetico-geometric sequence is

(iii) 1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, ……
Solution:
1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, …..
Here, 1, 2, 3, 4, 5, … are in A.P.
∴ a = 1, d = 1
∴ nth term = a + (n – 1)d
= 1 + (n – 1)1
= n

(iv) 3, 12, 36, 96, 240, ……
Solution:
3, 12, 36, 96, 240, ……
i.e., 1 × 3, 2 × 6, 3 × 12, 4 × 24, 5 × 48, …….
Here, 1, 2, 3, 4, 5, ….. are in A.P.
∴ nth term = n
Also, 3, 6, 12, 24, 48, ….. are in G.P.
∴ a = 3, r = 2
∴ nth term = ar^n-1 = 3 . (2^n-1)
∴ nth term of arithmetico-geometric sequence is

Question 2.
Find the sum to infinity of the following arithmetico-geometric sequence.
(i) \(1, \frac{2}{4}, \frac{3}{16}, \frac{4}{64}, \ldots\)
Solution:

(ii) \(3, \frac{6}{5}, \frac{9}{25}, \frac{12}{125}, \frac{15}{625}, \ldots\)
Solution:

∴ \(\frac{4}{5} S=\frac{15}{4}\)
∴ S = \(\frac{75}{16}\)

(iii) \(1, \frac{-4}{3}, \frac{7}{9}, \frac{-10}{27} \ldots\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.4

Question 1.
Verify whether the following sequences are H.P.
(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Solution:
\(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Here, the reciprocal sequence is 3, 5, 7, 9,…
t₁ = 3, t₂ = 5, t₃ = 7, t₄ = 9, …..
t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 2 = constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is a H.P.

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)
Solution:
\(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)
Here, the reciprocal sequence is 3, 6, 12, 24,…
t₁ = 3, t₂ = 6, t₃ = 12, ……
t₂ – t₁ = 3, t₃ – t₂ = 6
t₂ – t₁ ≠ t₃ – t₂
∴ The reciprocal sequence is not an A.P.
∴ The given sequence is not a H.P.

(iii) \(5, \frac{10}{17}, \frac{10}{32}, \frac{10}{47}, \ldots\)
Solution:

∴ The reciprocal sequence is an A.P.
∴ The given sequence is a H.P.

Question 2.
Find the nth term and hence find the 8th term of the following HPs.
(i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots\)
Solution:
\(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots\) are in H.P.
∴ 2, 5, 8, 11,… are in A.P.
∴ a = 2, d = 3
t_n = a + (n – 1)d
= 2 + (n – 1)(3)
= 3n – 1
∴ nth term of H.P. = \(\frac{1}{3 n-1}\)
∴ 8th term of H.P. = \(\frac{1}{3(8)-1}\) = \(\frac{1}{23}\)

(ii) \(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots\)
Solution:
\(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots\) are in H.P.
∴ 4, 6, 8, 10, … are in A.P.
∴ a = 4, d = 2
t_n = a + (n – 1)d
= 4 + (n – 1) (2)
= 2n + 2
∴ nth term of H.P. = \(\frac{1}{2 n+2}\)
∴ 8th term of H.P. = \(\frac{1}{2(8)+2}\) = \(\frac{1}{18}\)

(iii) \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \ldots\)
Solution:
\(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \ldots\) are in H.P.
∴ 5, 10, 15, 20, … are in A.P.
∴ a = 5, d = 5
t_n = a + (n – 1)d
= 5 + (n – 1) (5)
= 5n
∴ nth term of H.P. = \(\frac{1}{5 n}\)
∴ 8th term of H.P. = \(\frac{1}{5(8)}\) = \(\frac{1}{40}\)

Question 3.
Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \(\frac{16}{5}\) respectively.
Solution:
G.M. = 4, H.M. = \(\frac{16}{5}\)
Now, (G.M.)² = (A.M.) (H.M.)
∴ 4² = A.M. × \(\frac{16}{5}\)
∴ A.M. = 16 × \(\frac{5}{16}\)
∴ A.M. = 5

Question 4.
Find H.M. of two positive numbers whose A.M. and G.M. are \(\frac{15}{2}\) and 6.
Solution:
A.M. = \(\frac{15}{2}\), G.M. = 6
Now, (G.M.)² = (A.M.) (H.M.)
∴ 6² = \(\frac{15}{2}\) × H.M.
∴ H.M. = 36 × \(\frac{2}{15}\)
∴ H.M. = \(\frac{24}{5}\)

Question 5.
Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.
Solution:
A.M. = 75, H.M. = 48
Now, (G.M.)² = (A.M.) (H.M.)
∴ (G.M.)² = 75 × 48
∴ (G.M.)² = 25 × 3 × 16 × 3
∴ (G.M.)² = 5² × 4² × 3²
∴ G.M. = 5 × 4 × 3
∴ G.M. = 60

Question 6.
Insert two numbers between \(\frac{1}{4}\) and \(\frac{1}{3}\) so that the resulting sequence is a H.P.
Solution:
Let the required numbers be \(\frac{1}{\mathrm{H}_{1}}\) and \(\frac{1}{\mathrm{H}_{2}}\).
∴ \(\frac{1}{4}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{3}\) are in H.P.
∴ 4, H₁, H₂, 3 are in A.P.
t₁ = 4, t₂ = H₁, t₃ = H₂, t₄ = 3
∴ t₁ = a = 4, t₄ = 3
t_n = a + (n – 1)d
t₄ = 4 + (4 – 1)d
3 = 4 + 3d
3d = -1
∴ d = \(\frac{-1}{3}\)
H₁ = t₂ = a + d = 4 – \(\frac{1}{3}\) = \(\frac{11}{3}\)
H₂ = t₃ = a + 2d = 4 – \(\frac{2}{3}\) = \(\frac{10}{3}\)
∴ For resulting sequence to be H.P. we need to insert numbers \(\frac{3}{11}\) and \(\frac{3}{10}\).

Question 7.
Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.
Solution:
Let the required numbers be G₁ and G₂.
∴ 1, G₁, G₂, -27 are in G.P.
t₁ = 1, t₂ = G₁, t₃ = G₂, t₄ = -27
∴ t₁ = a = 1
t_n = ar^n-1
t₄ = (1) r^4-1
-27 = r³
r³ = (-3)³
∴ r = -3
∴ G₁ = t₂ = ar = 1(-3) = -3
G₂ = t₃ = ar² = 1(-3)² = 9
∴ For resulting sequence to be G.P. we need to insert numbers -3 and 9.

Question 8.
If the A.M. of two numbers exceeds their G.M. by 2 and their H.M. by \(\frac{18}{5}\), find the numbers.
Solution:
Let a and b be the two numbers.
A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)
According to the given conditions,

Consider, G = A – 2 = 10 – 2 = 8
\(\sqrt{a b}\) = 8
ab = 64
a(20 – a) = 64 …..[From (i)]
a² – 20a + 64 = 0
(a – 4)(a – 16) = 0
∴ a = 4 or a = 16
When a = 4, b = 20 – 4 = 16
When a = 16, b = 20 – 16 = 4
∴ The two numbers are 4 and 16.

Question 9.
Find two numbers whose A.M. exceeds their G.M. by \(\frac{1}{2}\) and their H.M. by \(\frac{25}{26}\).
Solution:
Let a and b be the two numbers.
A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)
According to the given conditions,

\(\sqrt{a b}\) = 6
ab = 36
a(13 – a) = 36 ……[From (i)]
a² – 13a + 36 = 0
(a – 4)(a – 9) = 0
∴ a = 4 or a = 9
When a = 4, b = 13 – 4 = 9
When a = 9, b = 13 – 9 = 4
∴ The two numbers are 4 and 9.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.3

Question 1.
Determine whether the sum to infinity of the following G.P.s exist, if exists find them.
(i) \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\)
Solution:

(ii) \(2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots\)
Solution:

(iii) \(-3,1, \frac{-1}{3}, \frac{1}{9}, \ldots\)
Solution:

(iv) \(\frac{1}{5}, \frac{-2}{5}, \frac{4}{5}, \frac{-8}{5}, \frac{16}{5}, \ldots\)
Solution:

(v) 9, 8.1, 7.29, ……
Solution:
9, 8.1, 7.29, …..
Here, a = 9, r = \(\frac{8.1}{9}\) = 0.9, |r| < 1
∴ Sum to infinity exists.
∴ Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)
= \(\frac{9}{1-0.9}\)
= \(\frac{9}{0.1}\)
= 90

Question 2.
Express the following recurring decimals as rational numbers.
(i) \(0 . \overline{7}\)
(ii) \(2 . \overline{4}\)
(iii) \(2.3 \overline{5}\)
(iv) \(51.0 \overline{2}\)
Solution:
(i) \(0 . \overline{7}\) = 0.7777… = 0.7 + 0.07 + 0.007 + ….
The terms 0.7, 0.07, 0.007,… are in G.P.
∴ a = 0.7, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(ii) \(2 . \overline{4}\) = 2.444 … = 2 + 0.4 + 0.04 + 0.004 + …
The terms 0.4, 0.04, 0.004,… are in G.P.
∴ a = 0.4, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = 10.11 < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(iii) \(2.3 \overline{5}\) = 2.3555… = 2.3 + 0.05 + 0.005 + 0.0005 + …
The terms 0.05,0.005,0.0005,… are in G.P.
∴ a = 0.05, r = \(\frac{0.005}{0.05}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(iv) \(51.0 \overline{2}\) = 51.0222 …. = 51 + 0.02 + 0.002 + 0.0002 + …..
The terms 0.02, 0.002, 0.0002,… are in G.P.
∴ a = 0.02, r = \(\frac{0.002}{0.02}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

Question 3.
If the common ratio of a G.P. is \(\frac{2}{3}\) and the sum to infinity is 12, find the first term.
Solution:
r = \(\frac{2}{3}\), sum to infinity = 12 ….. [Given]
Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)
12 = \(\frac{a}{1-\frac{2}{3}}\)
a = 12 × \(\frac{1}{3}\)
∴ a = 4

Question 4.
If the first term of the G.P. is 6 and its sum to infinity is \(\frac{96}{17}\), find the common ratio.
Solution:
a = 6, sum to infinity = \(\frac{96}{17}\) …..[Given]
Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)

Question 5.
The sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15, find the G.P.
Solution:
Let the required G.P. be a, ar, ar², ar³, …..
Sum to infinity of this G.P. = 5

Question 6.
Find
(i) \(\sum_{r=1}^{\infty} 4(0.5)^{r}\)
Solution:

(ii) \(\sum_{r=1}^{\infty}\left(-\frac{1}{3}\right)^{r}\)
Solution:

(iii) \(\sum_{r=0}^{\infty}(-8)\left(-\frac{1}{2}\right)^{r}\)
Solution:

(iv) \(\sum_{n=1}^{\infty} 0.4^{n}\)
Solution:

Question 7.
The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of
(i) the areas of all the squares.
(ii) the perimeters of all the squares.
Solution:

(i) Area of the 1st square = 12
Area of the 2nd square = \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)
Area of the 3rd square = \(\left(\frac{1}{2}\right)^{2}\)
and so on.
∴ Sum of the areas of all the squares

∴ Sum to infinity exists.
∴ Sum of the areas of all the squares = \(\frac{1}{1-\frac{1}{2}}\) = 2

(ii) Perimeter of 1st square = 4
Perimeter of 2nd square = 4\(\left(\frac{1}{\sqrt{2}}\right)\)
Perimeter of 3rd square = 4\(\left(\frac{1}{2}\right)\)
and so on.
Sum of the perimeters of all the squares

Question 8.
A ball is dropped from a height of 10 m. It bounces to a height of 6m, then 3.6 m, and so on. Find the total distance travelled by the ball.
Solution:
Here, on the first bounce, the ball will go 6 m and it will return 6 m.
On the second bounce, the ball will go 3.6 m and it will return 3.6 m, and so on.
Given that, a ball is dropped from a height of 10 m.
∴ Total distance travelled by the ball is = 10 + 2[6 + 3.6 + …]
The terms 6, 3.6 … are in G.P.
a = 6, r = 0.6, |r| = |0.6| < 1
∴ Sum to infinity exists.
∴ Total distance travelled by the ball

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

Question 1.
For the following G.P.s, find S_n.
(i) 3, 6, 12, 24, ……..
Solution:

(ii) p, q, \(\frac{\mathbf{q}^{2}}{\mathbf{p}}, \frac{\mathbf{q}^{3}}{\mathbf{p}^{2}}, \ldots\)
Solution:

(iii) 0.7, 0.07, 0.007, …….
Solution:

(iv) √5, -5, 5√5, -25, …….
Solution:

Question 2.
For a G.P.
(i) a = 2, r = \(-\frac{2}{3}\), find S₆.
Solution:

(ii) If S₅ = 1023, r = 4, find a.
Solution:

Question 3.
For a G.P.
(i) If a = 2, r = 3, S_n = 242, find n.
Solution:

(ii) For a G.P. sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.
Solution:

Question 4.
For a G.P.
(i) If t₃ = 20, t₆ = 160, find S₇.
Solution:

(ii) If t₄ = 16, t₉ = 512, find S₁₀.
Solution:

Question 5.
Find the sum to n terms
(i) 3 + 33 + 333 + 3333 + …..
Solution:
S_n = 3 + 33 + 333 +….. upto n terms
= 3(1 + 11 + 111 +….. upto n terms)
= \(\frac{3}{9}\)(9 + 99 + 999 + ….. upto n terms)
= \(\frac{3}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= \(\frac{3}{9}\)[(10 + 100 + 1000 + … upto nterms) – (1 + 1 + 1 + ….. n times)]
But 10, 100, 1000, ….. n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10

(ii) 8 + 88 + 888 + 8888 + …..
Solution:
S_n = 8 + 88 + 888 + … upto n terms
= 8(1 + 11 + 111 + … upto n terms)
= \(\frac{8}{9}\)(9 + 99 + 999 + … upto n terms)
= \(\frac{8}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= \(\frac{8}{9}\)[(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]
But 10, 100, 1000, … n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10

Question 6.
Find the sum to n terms
(i) 0.4 + 0.44 + 0.444 + …..
Solution:
S_n = 0.4 + 0.44 + 0.444 + ….. upto n terms
= 4(0.1 + 0.11 +0.111 + …. upto n terms)
= \(\frac{4}{9}\)(0.9 + 0.99 + 0.999 + … upto n terms)
= \(\frac{4}{9}\)[(1 – 0.1) + (1 – 0.01) + (1 – 0.001) … upto n terms]
= \(\frac{4}{9}\)[(1 + 1 + 1 + …n times) – (0.1 + 0.01 + 0.001 +… upto n terms)]
But 0.1, 0.01, 0.001, … n terms are in G.P. with
a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1

(ii) 0.7 + 0.77 + 0.777 + ……
Solution:
S_n = 0.7 + 0.77 + 0.777 + … upto n terms
= 7(0.1 + 0.11 + 0.111 + … upton terms)
= \(\frac{7}{9}\)(0.9 + 0.99 + 0.999 + … upto n terms)
= \(\frac{7}{9}\)[(1 – 0.1) + (1 – 0.01) + (1 – 0.001) +… upto n terms]
= \(\frac{7}{9}\)[(1 + 1 + 1 +… n times) – (0.1 + 0.01 + 0.001 +… upto n terms )]
But 0.1, 0.01, 0.001, … n terms are in G.P. with
a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1

Question 7.
Find the sum to n terms of the sequence
(i) 0.5, 0.05, 0.005, …..
Solution:

(ii) 0.2, 0.02, 0.002, ……
Solution:

Question 8.
For a sequence, if S_n = 2(3n – 1), find the nth term, hence showing that the sequence is a G.P.
Solution:

Question 9.
If S, P, R are the sum, product, and sum of the reciprocals of n terms of a G.P, respectively, then verify that \(\left[\frac{S}{R}\right]^{n}\) = P².
Solution:
Let a be the 1st term and r be the common ratio of the G.P.
∴ the G.P. is a, ar, ar², ar³, …, ar^n-1

Question 10.
If S_n, S_2n, S_3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that S_n(S_3n – S_2n) = (S_2n – S_n)².
Solution:
Let a and r be the 1st term and common ratio of the G.P. respectively.

Question 11.
Find
(i) \(\sum_{r=1}^{10}\left(3 \times 2^{r}\right)\)
Solution:

(ii) \(\sum_{r=1}^{10} 5 \times 3^{r}\)
Solution:

Question 12.
The value of a house appreciates 5% per year. How much is the house worth after 6 years if its current worth is Rs. 15 Lac. [Given: (1.05)⁵ = 1.28, (1.05)⁶ = 1.34]
Solution:
The value of a house is Rs. 15 Lac.
Appreciation rate = 5% = \(\frac{5}{100}\) = 0.05
Value of house after 1st year = 15(1 + 0.05) = 15(1.05)
Value of house after 6 years = 15(1.05) (1.05)⁵
= 15(1.05)⁶
= 15(1.34)
= 20.1 lac.

Question 13.
If one invests Rs. 10,000 in a bank at a rate of interest 8% per annum, how long does it take to double the money by compound interest? [(1.08)⁵ = 1.47]
Solution:
Amount invested = Rs. 10000
Interest rate = \(\frac{8}{100}\) = 0.08
amount after 1st year = 10000(1 + 0.08) = 10000(1.08)
Value of the amount after n years
= 10000(1.08) × (1.08)^n-1
= 10000(1.08)ⁿ
= 20000
∴ (1.08)ⁿ = 2
∴ (1.08)⁵ = 1.47 …..[Given]
∴ n = 10 years, (approximately)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.2

Question 1.
First, 6 faced die which is numbered 1 to 6 is thrown, then a 5 faced die which is numbered 1 to 5 is thrown. What is the probability that sum of the numbers on the upper faces of the dice is divisible by 2 or 3?
Solution:
When a 6 faced die and a 5 faced die are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4,4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(S) = 30
Let event A: The sum of the numbers on the upper faces of the dice is divisible by 2.
A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4)}
∴ n(A) = 15
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{15}{30}\)
Let event B: Sum of the numbers on the upper faces of the dice is divisible by 3.
B = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3)}
∴ n(B) = 10
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{10}{30}\)
Now,
A ∩ B = {(1, 5), (2,4), (3, 3), (4, 2), (5, 1)}
∴ n(A ∩ B) = 5
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{30}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{15}{30}+\frac{10}{30}-\frac{5}{30}\)
= \(\frac{20}{30}\)
= \(\frac{2}{3}\)

Question 2.
A card is drawn from a pack of 52 cards. What is the probability that,
(i) card is either red or black?
(ii) card is either black or a face card?
Solution:
One card can be drawn from the pack of 52 cards in \({ }^{52} \mathrm{C}_{1}\) = 52 ways.
∴ n(S) = 52
The pack of 52 cards consists of 26 red and 26 black cards.
(i) Let event A: A red card is drawn.
∴ Red card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26ways
∴ n(A) = 26
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{26}{52}\)
Let event B: A black card is drawn.
∴ Black card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26 ways.
∴ n(B) = 26
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{26}{52}\)
Since A and B are mutually exclusive events,
P(A ∩ B) = 0
∴ Required probability
P(A ∪ B) = P(A) + P(B)
= \(\frac{26}{52}+\frac{26}{52}\)
= 1

(ii) Let event A: A black card is drawn.
∴ Black card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26 ways.
n(A) = 26
n(A) 26 n(S) ~ 52
Let event B: A face card is drawn.
There are 12 face cards in the pack of 52 cards.
∴ 1 face card can be drawn in \({ }^{12} \mathrm{C}_{1}\) = 12 ways.
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}\)
There are 6 black face cards.
∴ n(A ∩ B) = 6
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{6}{52}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{26}{52}+\frac{12}{52}-\frac{6}{52}\)
= \(\frac{32}{52}\)
= \(\frac{8}{13}\)

Question 3.
A girl is preparing for National Level Entrance exam and State Level Entrance exam for professional courses. The chances of her cracking National Level exam is 0.42 and that of State Level exam is 0.54. The probability that she clears both the exams is 0.11. Find the probability that
(i) she cracks at least one of the two exams.
(ii) she cracks only one of the two.
(iii) she cracks none.
Solution:
Let event A: The girl cracks the National Level exam.
∴ P(A) = 0.42
Let event B: The girl cracks the State Level exam.
∴ P(B) = 0.54
Also, P(A ∩ B) = 0.11
(i) P(the girl cracks at least one of the two exams)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.42 + 0.54 – 0.11
= 0.85

(ii) P(the girl cracks only one of the two exams)
= P(A) – P(B) – 2P(A ∩ B)
= 0.42 + 0.54 – 2(0.11)
= 0.74

(iii) P(the girl cracks none of the exams)
= P(A’ ∩ B’)
= P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – 0.85
= 0.15

Question 4.
A bag contains 75 tickets numbered from 1 to 75. One ticket is drawn at random. Find the probability that,
(i) number on the ticket is a perfect square or divisible by 4.
(ii) number on the ticket is a prime number or greater than 40.
Solution:
Out of the 75 tickets, one ticket can be drawn in \({ }^{75} \mathrm{C}_{1}\) = 75 ways.
∴ n(S) = 75
(i) Let event A: The number on the ticket is a perfect square.
∴ A = {1, 4, 9, 16, 25, 36, 49, 64}
∴ n(A) = 8
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{8}{75}\)
Let event B: The number on the ticket is divisible by 4.
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72}
∴ n(B) = 18
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{18}{75}\)
Now, A ∩ B = {4, 16, 36, 64}
∴ n(A ∩ B) = 4
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{4}{75}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{8}{75}+\frac{18}{75}-\frac{4}{75}\)
= \(\frac{22}{75}\)

(ii) Let event A: The number on the ticket is a prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(A) = 21
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{21}{75}\)
Let event B: The number is greater than 40.
∴ B = {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75}
∴ n(B) = 35
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{35}{75}\)
Now,
A ∩ B = {41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(A ∩ B) = 9
∴ n(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{9}{75}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{21}{75}+\frac{35}{75}-\frac{9}{75}\)
= \(\frac{47}{75}\)

Question 5.
The probability that a student will pass in French is 0.64, will pass in Sociology is 0.45 and will pass in both is 0.40. What is the probability that the student will pass in at least one of the two subjects?
Solution:
Let event A: The student will pass in French.
∴ P(A) = 0.64
Let event B: The student will pass in Sociology.
∴ P(B) = 0.45
Also, P(A ∩ B) = 0.40
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.64 + 0.45 – 0.40
= 0.69

Question 6.
Two fair dice are thrown. Find the probability that the number on the upper face of the first die is 3 or sum of the numbers on their upper faces is 6.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let event A: The number on the upper face of the first die is 3.
∴ A = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}
∴ n(A) = 6
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{36}\)
Let event B: Sum of the numbers on their upper faces is 6.
∴ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
∴ n(B) = 5
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{36}\)
Now, A ∩ B = {(3, 3)}
∴ n(A ∩ B) = 1
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{1}{36}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{6}{36}+\frac{5}{36}-\frac{1}{36}\)
= \(\frac{10}{36}\)
= \(\frac{5}{18}\)

Question 7.
For two events A and B of a sample space S, if P(A) = \(\frac{3}{8}\), P(B) = \(\frac{1}{2}\) and P(A ∪ B) = \(\frac{5}{8}\). Find the value of the following.
(a) P(A ∩ B)
(b) P(A’ ∩ B’)
(c) P(A’ ∪ B’)
Solution:
Here, P(A) = \(\frac{3}{8}\), P(B) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{5}{8}\)
(a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{3}{8}+\frac{1}{2}-\frac{5}{8}\)
= \(\frac{1}{4}\)

(b) P(A’ ∩ B’) = P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – \(\frac{5}{8}\)
= \(\frac{3}{8}\)

(c) P(A’ ∪ B’) = P(A ∩ B)’
= 1 – P(A ∩ B)
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)

Question 8.
For two events A and B of a sample space S, if P(A ∪ B) = \(\frac{5}{6}\), P(A ∩ B) = \(\frac{1}{3}\) and P(B’) = \(\frac{1}{3}\), then find P(A).
Solution:
Here, P(A ∪ B) = \(\frac{5}{6}\), P(A ∩ B) = \(\frac{1}{3}\), P(B’) = \(\frac{1}{3}\)
P(B) = 1 – P(B’)
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
Since P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(\frac{5}{6}\) = P(A) + \(\frac{2}{3}-\frac{1}{3}\)
∴ \(\frac{5}{6}\) = P(A) + \(\frac{1}{3}\)
∴ P(A) = \(\frac{5}{6}-\frac{1}{3}\) = \(\frac{1}{2}\)

Question 9.
A bag contains 5 red, 4 blue and an unknown number m of green balls. If the probability of getting both the balls green, when two balls are selected at random is \(\frac{1}{7}\), find m.
Solution:
Total number of balls in the bag = 5 + 4 + m = 9 + m
Two balls are selected from (9 + m) balls in \({ }^{9+m} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{9+m} \mathrm{C}_{2}\)
Let event A: The two balls selected are green.
∴ 2 balls can be selected from m balls in \({ }^{\mathrm{m}} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{\mathrm{m}} \mathrm{C}_{2}\)

(9 + m)(8 + m) = 7m(m – 1)
72 + 9m + 8m + m2 = 7m2 – 7m
6m2 – 24m – 72 = 0
m2 – 4m – 12 = 0
(m – 6)(m + 2) = 0
m = 6 or m = -2
Since number of balls cannot be negative, m ≠ -2
∴ m = 6

Question 10.
From a group of 4 men, 4 women and 3 children, 4 persons are selected at random. Find the probability that,
(i) no child is selected.
(ii) exactly 2 men are selected.
Solution:
The group consists of 4 men, 4 women and 3 children, i.e., 4 + 4 + 3 = 11 persons.
4 persons are to be selected from this group.
∴ 4 persons can be selected from 11 persons in \({ }^{11} \mathrm{C}_{4}\) ways.
∴ n(S) = \({ }^{11} \mathrm{C}_{4}\)
(i) Let event A: No child is selected.
∴ 4 persons can be selected from 4 men and 4 women, i.e., from 8 persons in \({ }^{8} \mathrm{C}_{4}\) ways.
∴ n(A) = \({ }^{8} \mathrm{C}_{4}\)

(ii) Let event B: Exactly 2 men are selected.
∴ 2 men are selected from 4 men in \({ }^{4} \mathrm{C}_{2}\) ways, and remaining 2 persons are selected from 7 persons (i.e., 4 women and 3 children) in \({ }^{7} \mathrm{C}_{2}\) ways.

Question 11.
A number is drawn at random from the numbers 1 to 50. Find the probability that it is divisible by 2 or 3 or 10.
Solution:
One number can be drawn at random from the numbers 1 to 50 in \({ }^{50} \mathrm{C}_{1}\) = 50 ways.
∴ n(S) = 50
Let event A: The number drawn is divisible by 2.
∴ A = {2, 4, 6, 8, 10, …, 48, 50}
∴ n(A) = 25
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{25}{50}\)
Let event B: The number drawn is divisible by 3.
B = {3, 6, 9, 12, …, 48}
∴ n(B) = 16
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{16}{50}\)
Let event C: The number drawn is divisible by 10.
C = {10, 20, 30, 40, 50}
∴ n(C) = 5
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{5}{50}\)
Now, A ∩ B = {6, 12, 18, 24, 30, 36, 42, 48}
∴ n(A ∩ B) = 8
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{8}{50}\)
B ∩ C = {30}
∴ n(B ∩ C) = 1
∴ P(B ∩ C) = \(\frac{\mathrm{n}(\mathrm{B} \cap \mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{1}{50}\)
A ∩ C = {10, 20, 30, 40, 50}
∴ n(A ∩ C) = 5
∴ P(A ∩ C) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{5}{50}\)
A ∩ B ∩ C = {30}
∴ n(A ∩ B ∩ C) = 1
∴ P(A ∩ B ∩C) = \(\frac{n(A \cap B \cap C)}{n(S)}=\frac{1}{50}\)
∴ P(the number is divisible by 2 or 3 or 10)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)
= \(\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{8}{50}-\frac{1}{50}-\frac{5}{50}+\frac{1}{50}\)
= \(\frac{33}{50}\)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.1

Question 1.
There are four pens: Red, Green, Blue, and Purple in a desk drawer of which two pens are selected at random one after the other with replacement. State the sample space and the following events.
(a) A : Select at least one red pen.
(b) B : Two pens of the same colour are not selected.
Solution:
The drawer contains 4 pens out of which one is red (R), one is green (G), one is blue (B) and the other one is purple (P).
From this drawer, two pens are selected one after the other with replacement.
∴ The sample space S is given by
S = {RR, RG, RB, RP, GR, GG, GB, GP, BR, BG, BB, BP, PR, PG, PB, PP}
(a) A : Select at least one red pen.
At least one means one or more than one.
∴ A = {RR, RG, RB, RP, GR, BR, PR}

(b) B : Two pens of the same colour are not selected.
B = {RG, RB, RP, GR, GB, GP, BR, BG, BP, PR, PG, PB}

Question 2.
A coin and a die are tossed simultaneously. Enumerate the sample space and the following events.
(a) A : Getting a tail and an odd number.
(b) B : Getting a prime number.
(c) C : Getting head and a perfect square.
Solution:
When a coin and a die are tossed simultaneously, the sample space S is given by
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
(a) A : Getting a tail and an odd number.
∴ A = {(T, 1), (T, 3), (T, 5)}

(b) B : Getting a prime number.
∴ B = {(H, 2), (H, 3), (H, 5), (T, 2), (T, 3), (T, 5)}

(c) C : Getting a head and a perfect square.
∴ C = {(H, 1), (H, 4)}

Question 3.
Find n(S) for each of the following random experiments.
(a) From an urn containing 5 gold and 3 silver coins, 3 coins are drawn at random.
(b) 5 letters are to be placed into 5 envelopes such that no envelope is empty.
(c) 6 books of different subjects are arranged on a shelf.
(d) 3 tickets are drawn from a box containing 20 lottery tickets.
Solution:
(a) There are 5 gold and 3 silver coins, i.e., 8 coins.
3 coins can be drawn from these 8 coins in \({ }^{8} \mathrm{C}_{3}\) ways.
∴ n(s) = \({ }^{8} \mathrm{C}_{3}=\frac{8 !}{5 ! 3 !}=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}=56\)

(b) 5 letters have to be placed in 5 envelopes in such a way that no envelope is empty.
∴ The first letter can be placed into 5 envelopes in 5 different ways, the second letter in 4 ways.
Similarly, the third, fourth and fifth letters can be placed in 3 ways, 2 ways and 1 way, respectively.
∴ Total number of ways = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(c) 6 books can be arranged on a shelf in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(d) 3 tickets are drawn at random from 20 tickets.
∴ 3 tickets can be selected in \({ }^{20} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{3}=\frac{20 !}{17 ! 3 !}=\frac{20 \times 19 \times 18 \times 17 !}{17 ! \times 3 \times 2 \times 1}\) = 1140

Question 4.
Two fair dice are thrown. State the sample space and write the favourable outcomes for the following events.
(a) A : Sum of numbers on two dice is divisible by 3 or 4.
(b) B : The sum of numbers on two dice is 7.
(c) C : Odd number on the first die.
(d) D : Even number on the first die.
(e) Check whether events A and B are mutually exclusive and exhaustive.
(f) Check whether events C and D are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) A: Sum of the numbers on two dice is divisible by 3 or 4.
∴ A = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}

(b) B: Sum of the numbers on two dice is 7.
∴ B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(c) C: Odd number on the first die.
∴ C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(d) D: Even number on the first die.
∴ D = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(e) A and B are mutually exclusive events as A ∩ B = Φ.
A ∪ B = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S
∴ A and B are not exhaustive events as A ∪ B ≠ S.

(f) C and D are mutually exclusive events as C ∩ D = Φ.
C ∪ D = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S
∴ C and D are exhaustive events.

Question 5.
A bag contains four cards marked as 5, 6, 7, and 8. Find the sample space if two cards are drawn at random
(a) with replacement.
(b) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8. Two cards are to be drawn from this bag.
(a) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6, 8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}

(b) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}

Question 6.
A fair die is thrown two times. Find the probability that
(a) the sum of the numbers on them is 5.
(b) the sum of the numbers on them is at least 8.
(c) the first throw gives a multiple of 2 and the second throw gives a multiple of 3.
(d) product of numbers on them is 12.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) Let event A: Sum of the numbers on uppermost face is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

(b) Let event B: Sum of the numbers on uppermost face is at least 8 (i.e., 8 or more than 8)
∴ B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 15
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)

(c) Let event C: First throw gives a multiple of 2 and second throw gives a multiple of 3.
∴ C = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)}
∴ n(C) = 6
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(d) Let event D: The product of the numbers on uppermost face is 12.
∴ D = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(D) = 4
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

Question 7.
Two cards are drawn from a pack of 52 cards. Find the probability that
(a) one is a face card and the other is an ace card.
(b) one is a club and the other is a diamond.
(c) both are from the same suit.
(d) both are red cards.
(e) one is a heart card and the other is a non-heart card.
Solution:
Two cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{2}\)

(a) Let event A: Out of the two cards drawn, one is a face card and the other is an ace card.
There are 12 face cards and 4 ace cards in a pack of 52 cards.
∴ One face card can be drawn from 12 face cards in \({ }^{12} \mathrm{C}_{1}\) ways and one ace card can be drawn from 4 ace cards in \({ }^{4} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{12 \times 4}{\frac{52 \times 51}{2 \times 1}}=\frac{8}{221}\)

(b) Let event B: Out of the two cards drawn, one is club and the other is a diamond card.
There are 13 club cards and 13 diamond cards.
∴ One club card can be drawn from 13 club cards in \({ }^{13} \mathrm{C}_{1}\) ways and one diamond card can be drawn from 13 diamond cards in \({ }^{13} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{{ }^{13} C_{1} \times{ }^{13} C_{1}}{{ }^{52} C_{2}}=\frac{13 \times 13}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{102}\)

(c) Let event C: Both the cards drawn are of the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 2 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{2}\) ways.
∴ n(C) = \({ }^{13} \mathrm{C}_{2} \times 4\)
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 13 \times 12}{52 \times 51}=\frac{4}{17}\)

(d) Let event D: Both the cards drawn are red.
There are 26 red cards in the pack of 52 cards.
∴ 2 cards can be drawn from them in \({ }^{26} \mathrm{C}_{2}\) ways.
∴ n(D) = \({ }^{26} \mathrm{C}_{2}\)
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{{ }^{26} C_{2}}{{ }^{52} C_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)

(e) Let event E: Out of the two cards drawn, one is heart and other is non-heart.
There are 13 heart cards in a pack of 52 cards, i.e., 39 cards are non-heart.
∴ One heart card can be drawn from 13 hdart cards in \({ }^{13} \mathrm{C}_{1}\) ways and one non-heart card can be drawn from 39 cards in \({ }^{39} \mathrm{C}_{1}\) ways.
∴ n(E) = \({ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}\)
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{13 \times 39}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{34}\)

Question 8.
Three cards are drawn from a pack of 52 cards. Find the chance that
(a) two are queen cards and one is an ace card.
(b) at least one is a diamond card.
(c) all are from the same suit.
(d) they are a king, a queen, and a jack.
Solution:
3 cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{3}\)

(a) Let event A: Out of the three cards drawn, 2 are queens and 1 is an ace card.
There are 4 queens and 4 aces in a pack of 52 cards.
∴ 2 queens can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{2}\) ways and 1 ace can be drawn out of 4 aces in \({ }^{4} \mathrm{C}_{1}\) ways.

(b) Let event B: Out of the three cards drawn, at least one is a diamond.
∴ B’ is the event that all 3 cards drawn are non-diamond cards.
In a pack of 52 cards, there are 39 non-diamond cards.
∴ 3 non-diamond cards can be drawn in \({ }^{39} \mathrm{C}_{3}\) ways.

(c) Let event C: All the cards drawn are from the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 3 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{3}\) ways.

(d) Let event D: The cards drawn are a king, a queen, and a jack.
There are 4 kings, 4 queens and 4 jacks in a pack of 52 cards.
∴ 1 king can be drawn from 4 kings in \({ }^{4} \mathrm{C}_{1}\) ways,
1 queen can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{1}\) ways and
1 jack can be drawn from 4 jacks in \({ }^{4} \mathrm{C}_{1}\) ways.

Question 9.
From a bag containing 10 red, 4 blue, and 6 black balls, a ball is drawn at random. Find the probability of drawing
(a) a red bail.
(b) a blue or black ball.
(c) not a black ball.
Solution:
The bag contains 10 red, 4 blue, and 6 black balls,
i.e., 10 + 4 + 6 = 20 balls.
One ball can be drawn from 20 balls in \({ }^{20} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{1}\) = 20

(a) Let event A: Ball drawn is red.
There are total 10 red balls.
∴ 1 red ball can be drawn from 10 red balls in \({ }^{10} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{10} \mathrm{C}_{1}\) = 10
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(b) Let event B: The ball drawn is blue or black.
There are 4 blue and 6 black balls.
∴ 1 blue ball can be drawn from 4 blue balls in \({ }^{4} \mathrm{C}_{1}\) ways
or 1 black ball can be drawn from 6 black balls in \({ }^{6} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{4} \mathrm{C}_{1}\) + \({ }^{6} \mathrm{C}_{1}\) = 4 + 6 = 10
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(c) Let event C: Ball drawn is not black,
i.e., ball drawn is red or blue.
There are total 14 red and blue balls.
∴ 1 ball can be drawn from 14 balls in \({ }^{14} \mathrm{C}_{1}\) ways.
∴ n(C) = \({ }^{14} \mathrm{C}_{1}\) = 14
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{14}{20}=\frac{7}{10}\)

Question 10.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. Find the probability that,
(a) number on the ticket is divisible by 6.
(b) the number on the ticket is a perfect square.
(c) the number on the ticket is prime.
(d) the number on the ticket is divisible by 3 and 5.
Solution:
The box contains 75 tickets numbered 1 to 75.
∴ 1 ticket can be drawn from the box in \({ }^{75} \mathrm{C}_{1}\) = 75 ways.
∴ n(S) = 75

(a) Let event A: Number on the ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{75}=\frac{4}{25}\)

(b) Let event B: Number on the ticket is a perfect square.
∴ B = (1, 4, 9, 16, 25, 36, 49, 64}
∴ n(B) = 8
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{8}{75}\)

(c) Let event C: Number on the ticket is a prime number.
∴C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}

(d) Let event D: Number on the ticket is divisible by 3 and 5,
i.e., divisible by L.C.M. of 3 and 5,
i.e., 15.
∴D = {15, 30, 45, 60, 75}
∴ n(D) = 5
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{5}{75}=\frac{1}{15}\)

Question 11.
What is the chance that a leap year, selected at random, will contain 53 Sundays?
Solution:
A leap year consists of 366 days.
It has 52 complete weeks and two more days.
These two days can be {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun)}.
∴ n(S) = 7
Let event E : There are 53 Sundays.
∴ E = {(Sun, Mon), (Sat, Sun)}
∴ n(E) = 2
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{2}{7}\)

Question 12.
Find the probability of getting both red balls, when from a bag containing 5 red and 4 black balls, two balls are drawn,
(i) with replacement
(ii) without replacement
Solution:
The bag contains 5 red and 4 black balls,
i.e., 5 + 4 = 9 balls.
(i) 2 balls can be drawn from 9 balls with replacement in \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) = 9 × 9 = 81
Let event A: Balls drawn are red.
2 red balls can be drawn from 5 red balls with replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) = 5 × 5 = 25
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{25}{81}\)

(ii) 2 balls can be drawn from 9 balls without replacement in \({ }^{9} C_{1} \times{ }^{8} C_{1}\) ways.
∴ n(S) = \({ }^{9} C_{1} \times{ }^{8} C_{1}\) = 9 × 8 = 72
2 red balls can be drawn from 5 red balls without replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) = 5 × 4 = 20
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{20}{72}=\frac{5}{18}\)

Question 13.
A room has three sockets for lamps. From a collection of 10 bulbs of which 6 are defective. At night a person selects 3 bulbs, at random and puts them in sockets. What is the probability that
(i) room is still dark.
(ii) the room is lit.
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 bulbs in \({ }^{10} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{10} \mathrm{C}_{3}\)

(i) Let event A: The room is dark.
For event A to happen the bulbs should be selected from the 6 defective bulbs. This can be done in \({ }^{6} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{6} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{6} C_{3}}{{ }^{10} C_{3}}=\frac{6 \times 5 \times 4}{10 \times 9 \times 8}=\frac{1}{6}\)

(ii) Let event A’: The room is lit.
∴ P(Room is lit) = 1 – P(Room is not lit)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)

Question 14.
Letters of the word MOTHER are arranged at random. Find the probability that in the arrangement
(a) vowels are always together.
(b) vowels are never together.
(c) O is at the beginning and end with T.
(d) starting with a vowel and ending with a consonant.
Solution:
There are 6 letters in the word MOTHER.
These letters can be arranged among themselves in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6!
(a) Let event A: Vowels are always together.
The word MOTHER consists of 2 vowels (O, E) and 4 consonants (M, T, H, R).
2 vowels can be arranged among themselves in \({ }^{2} \mathbf{P}_{2}\) = 2! ways.
Let us consider 2 vowels as one group.
This one group with 4 consonants can be arranged in \({ }^{5} \mathrm{P}_{5}\) = 5! ways.
∴ n(A) = 2! × 5!
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2 ! \times 5 !}{6 !}=\frac{1}{3}\)

(b) Let event B: Vowels are never together.
4 consonants create 5 gaps, in which vowels are arranged.
Consider the following arrangement of consonants
_C_C_C_C_
2 vowels can be arranged in 5 gaps in \({ }^{5} \mathbf{P}_{2}\) ways.
Also 4 consonants can be arranged among themselves in \({ }^{4} \mathbf{P}_{4}\) = 4! ways.
∴ n(B) = 4! × \({ }^{5} \mathbf{P}_{2}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4 ! \times{ }^{5} P_{2}}{6 !}=\frac{4 ! \times 5 \times 4}{6 \times 5 \times 4 !}=\frac{2}{3}\)

(iii) Let event C: Word begin with O and end with T.
Thus first and last letters can be arranged in one way each and the remaining 4 letters can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways
∴ n(C) = 4! × 1 × 1 = 4!
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{4 !}{6 !}=\frac{1}{30}\)

(d) Let event D: Word starts with a vowel and ends with a consonant.
There are 2 vowels and 4 consonants in the word MOTHER.
∴ The first place can be arranged in 2 different ways and the last place can be arranged in 4 different ways.
Now, the remaining 4 letters (3 consonants and 1 vowel) can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways.
∴ n(D) = 2 × 4 × 4!
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{2 \times 4 \times 4 !}{6 !}=\frac{4}{15}\)

Question 15.
4 letters are to be posted in 4 post boxes. If any number of letters can be posted in any of the 4 post boxes, what is the probability that each box contains only one letter?
Solution:
There are 4 letters and 4 post boxes.
Since any number of letters can be posted in all 4 post boxes,
so each letter can be posted in different ways.
∴ n(S) = 4 × 4 × 4 × 4
Let event A: Each box contains only one letter.
∴ 1st letter can be posted in 4 different ways.
Since each box contains only one letter, 2nd letter can be posted in 3 different ways.
Similarly, 3rd and 4th letters can be posted in 2 different ways and 1 way respectively.
∴ n(A) = 4 × 3 × 2 × 1
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4 \times 3 \times 2 \times 1}{4 \times 4 \times 4 \times 4}=\frac{3}{32}\)

Question 16.
15 professors have been invited for a round table conference by the Vice-chancellor of a university. What is the probability that two particular professors occupy the seats on either side of the Vice-chancellor during the conference?
Solution:
Since a Vice-chancellor invited 15 professors for a round table conference, there were all 16 persons in the conference.
These 16 persons can be arranged among themselves around a round table in (16 – 1)! = 15! ways.
∴ n(S) = 15!
Let event A: Two particular professors be seated on either side of the Vice-chancellor.
Those two particular persons sit on either side of a Vice chancellor in \({ }^{2} \mathrm{P}_{2}\) = 2! ways.
Thus the remaining 13 persons can be arranged in \({ }^{13} \mathrm{P}_{13}\) = 13! ways.
∴ n(A) = 13! 2!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{13 ! \times 2 !}{15 !}=\frac{13 ! \times 2 \times 1}{15 \times 14 \times 13 !}=\frac{1}{105}\)

Question 17.
A bag contains 7 black and 4 red balls. If 3 balls are drawn at random, find the probability that
(i) all are black.
(ii) one is black and two are red.
Solution:
The bag contains 7 black and 4 red balls,
i.e., 7 + 4 = 11 balls.
∴ 3 balls can be drawn out of 11 balls in \({ }^{11} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{11} \mathrm{C}_{3}\)
(i) Let event A: All 3 balls drawn are black.
There are 7 black balls.
∴ 3 black balls can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{7} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{3}}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33}\)

(ii) Let event B: Out of 3 balls drawn, one is black and two are red.
There are 7 black and 4 red balls.
∴ One black ball can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{1}\) ways and 2 red balls can be drawn from 4 red balls in \({ }^{4} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{7} C_{1} \times{ }^{4} C_{2}}{{ }^{11} C_{3}}=\frac{7 \times \frac{4 \times 3}{2}}{\frac{11 \times 10 \times 9}{3 \times 2 \times 1}}=\frac{14}{55}\)