Class 10

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.2 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.2 Chapter 6 Statistics Questions With Answers Maharashtra Board

Statistics Practice Set 6.2 Question 1.
The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.

Solution:

Cumulative frequency which is just greater than (or equal) to 500 is 650.
∴ The median class is 10 – 12.
Now, L = 10, f = 500, cf = 150, h = 2

∴ The median of the number of hours the workers work is 11.4 hours.

10th Class Algebra Practice Set 6.2 Question 2.
The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.

Solution:

Here, total frequency = ∑f_i = N = 250
∴ \(\frac { N }{ 2 } \) = \(\frac { 250 }{ 2 } \) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 153.
∴ The median class is 150 – 200.
Now, L = 150, f = 90, cf = 63, h = 50

∴ The median of the given data is 184 mangoes (approx).

Statistics Class 10 Practice Set 6.2 Question 3.
The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.

Solution:

Here, total frequency = ∑f_i = N = 200
∴ \(\frac { N }{ 2 } \) = \(\frac { 200 }{ 2 } \) = 100
Cumulative frequency which is just greater than (or equal) to 100 is 184.
∴ The median class is 74.5 – 79.5.
Now, L = 74.5, f = 85, cf = 99, h = 5

∴ The median of the given data is 75 km/hr (approx.).

Practice Set 6.2 Geometry Class 10 Question 4.
The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.

Solution:

Cumulative frequency which is just greater than (or equal) to 52.5 is 67.
∴ The median class is 50 – 60.
Now, L = 50, f = 20, cf = 47, h = 10

∴ The median of the productions is 52750 bulbs (approx.).

Practice Set 6.2 Question 1.
If the number of scores is odd, then the (\(\frac { n+1 }{ 2 } \))^th score is the median of the data. That is, the number of scores below as well as above \({ K }_{ \frac { n+1 }{ 2 } }\) is \(\frac { n-1 }{ 2 } \) Verify the fact by taking n = 2m + I. (Textbk pg. no. 139)
Solution:

The sequence of the terms of scores is 1,2, 3, …….., m, m + 1, m + 2, …, 2m + 1
Thus, we have to prove that m + 1 is the middle term if the number of scores is 2m + 1
i.e. to prove
number of terms from 1 to m = number of terms from m + 2 to 2m + 1 …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1,d = 1, t_n1 = m
t_n1 = a + (n₁ – 1) d
∴ m = 1 + (n₁ – 1)1
∴ m = 1 + n₁ – 1
∴ m = n₁
Consider the R.H.S. of equation (ii)
The sequence is an A.P. with a = m + 2, d = 1, t_n2 = 2m + 1
t_n2 = a + (n₂ – 1)d
∴ 2m + 1 = m + 2 + (n₂ – 1)1
∴ 2m + 1 = m + n₂ + 1
∴ m = n₂
∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = \(\frac { n-1 }{ 2 } \)
∴ m + 1 is the middle term if the number of scores is 2m + 1.

Question 2.
If the number of the scores is even, then the mean of the middle two terms is the median. This is because the number of terms below \({ K }_{ \frac { n }{ 2 } }\) and above \({ K }_{ \frac { n+2 }{ 2 } }\) is equal, which is \(\frac { n-2 }{ 2 } \). Verify this by taking n = 2m. (Textbook pg. no. 139)
Solution:

The sequence of the terms of scores is 1, 2, 3 … m – 1, m, m + 1, m + 2,…., 2m
Thus, we have to prove that m and m + 1 are the middlemost terms if the number of scores is 2m.
i.e. to prove
number of terms from 1 to m – 1 = number of terms from m + 2 to 2m …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1, d = 1, t_n1 = m – 1
t_n1 = a + (n₁ – 1)d
∴ m – 1 = 1 + (n₁ – 1)1
∴m – 1 = 1 + n₁ – 1
∴ n₁ = m – 1
Consider the R.H.S. of equation (i)
The sequence is an A.P. with a = m + 2, d = 1, t_n2= 2m
t_n2= a + (n₂ – 1) d
∴ 2m = m + 2 + (n₂ – 1)1
∴ 2m = m + 2 + n₂ – 1
∴ n₂ = m – 1
∴ number of terms from 1 to m – 1 = number of terms from m + 2 to 2m = m – 1 = \(\frac { n-2 }{ 2 } \)
∴ m and m + 1 are the middlemost terms if the number of scores is 2m.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

10th Standard Maths 1 Problem Set 5 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Problem Set 5 Chapter 5 Probability Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for each of the following questions.

i. Which number cannot represent a probability?
(A) \(\frac { 2 }{ 3 } \)
(B) 1.5
(C) 15%
(D) 0.7
Answer:
The probability of any 0 to 1 or 0% to 100%. event is from
(B)

ii. A die is rolled. What is the probability that the number appearing on upper face is less than 3?
(A) \(\frac { 1 }{ 6 } \)
(B) \(\frac { 1 }{ 3 } \)
(C) \(\frac { 1 }{ 2 } \)
(D) 0
Answer:
(B)

iii. What is the probability of the event that a number chosen from 1 to 100 is a prime number?
(A) \(\frac { 1 }{ 5 } \)
(B) \(\frac { 6 }{ 25 } \)
(C) \(\frac { 1 }{ 4 } \)
(D) \(\frac { 13 }{ 50 } \)
Answer:
n(S) = 100
Let A be the event that the number chosen 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, 79, 83, 89, 97}
∴ n(A) = 25
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 25 }{ 100 } \) = \(\frac { 1 }{ 4 } \)
(C)

iv. There are 40 cards in a bag. Each bears a number from 1 to 40. One card is drawn at random. What is the probability that the card bears a number which is a multiple of 5?
(A) \(\frac { 1 }{ 5 } \)
(B) \(\frac { 3 }{ 5 } \)
(C) \(\frac { 4 }{ 5 } \)
(D) \(\frac { 1 }{ 3 } \)
Answer:
(A)

v. If n(A) = 2, P(A) = \(\frac { 1 }{ 5 } \), then n(S) = ?
(A) 10
(B) \(\frac { 5 }{ 2 } \)
(C) \(\frac { 2 }{ 5 } \)
(D) \(\frac { 1 }{ 3 } \)
Answer:
(A)

Question 2.
Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are \(\frac { 4 }{ 5 } \), 0.83 and 58% respectively. Who had the greatest probability of success ?
Solution:
The probability that the ball is dropped in the basket by John = \(\frac { 4 }{ 5 } \) = 0.80
The probability that the ball is dropped in the basket by Vasim = 0.83
The probability that the ball is dropped in the basket by Akash = 58% = \(\frac { 58 }{ 100 } \) = 0.58
0.83 > 0.80 > 0.58
∴ Vasim has the greatest probability of success.

Question 3.
In a hockey team there are 6 defenders , 4 offenders and 1 goalie. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that:
i. The goalie will be selected.
ii. A defender will be selected.
Solution:
Total number of players in the hockey team
= 6 + 4 + 1 = 11
∴ n(S) = 11

i. Let A be the event that the captain selected will be a goalie.
There is only one goalie in the hockey team.

ii. Let B be the event that the captain selected will be a defender.
There are 6 defenders in the hockey team.

Question 4.
Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card ?
Solution:
Each card bears an English alphabet.
∴ n(S) = 26
Let A be the event that the card drawn is a vowel card.
There are 5 vowels in English alphabets.
∴ A = {a, e, i, o, u}
∴ n(A) = 5

∴ The probability that the card drawn is a vowel card is \(\frac { 5 }{ 26 } \).

Question 5.
A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
i. a red balloon.
ii. a blue balloon,
iii. a green balloon.
Solution:
Let the 2 red balloon be R₁, R₂,
3 blue balloons be B₁, B₂, B₃, and
4 green balloons be G₁, G₂, G₃, G₄.
∴ Sample space
S = {R₁, R₂, B₁, B₂, B₃, G₁, G₂, G₃, G₄}
∴ n(S) = 9

i. Let A be the event that Pranali gets a red balloon.

∴ The probability that Pranali gets a red balloon is \(\frac { 2 }{ 9 } \)

ii. Let B be the event that Pranali gets a blue balloon.

∴ The probability that Pranali gets a blue balloon is \(\frac { 1 }{ 3 } \).

iii. Let C be the event that Pranali gets a green balloon.

∴ The probability that Pranali gets a green balloon is \(\frac { 4 }{ 9 } \).

Question 6.
A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?
Solution:
Let 5 red pens be R₁, R₂, R₃, R₄, R₅.
8 blue pens be B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈. and
3 green pens be G₁, G₂, G₃.
∴ Sample space
S = {R₁, R₂, R₃, R₄, R₅, B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈, G₁, G₂, G₃}
∴ n(S) = 16
Let A be the event that Rutuja picks a blue pen.
∴ A = {B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈}

∴ The probability that Rutuja picks a blue pen is \(\frac { 1 }{ 2 } \).

Question 7.
Six faces of a die are as shown below.

If the die is rolled once, find the probability of
i. ‘A’ appears on upper face.
ii. ‘D’ appears on upper face.
Solution:
Sample space
S = {A, B, C, D, E, A}
∴ n (S) = 6
i. Let R be the event that ‘A’ appears on the upper face.
∴ R = {A, A}
∴ n(R) = 2

ii. Let Q be the event that ‘D’ appears on the upper face.
Total number of faces having ‘D’ on it = 1
Q = {D}
∴ n(Q) = 1

Question 8.
A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
i. an odd number.
ii. a complete square number.
Solution:
Sample space,
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
∴ n(S) = 30

i. Let A be the event that the ticket drawn bears an odd number.
∴ A = {1,3,5,7,9,11,13,15,17,19,21, 23,25,27,29}
∴ n(A) =15

ii. Let B be the event that the ticket drawn bears a complete square number.
∴ B = {1,4,9,16,25}
∴ n(B) = 5

Question 9.
Length and breadth of a rectangular garden are 77 m and 50 m. There is a circular lake in the garden having diameter 14 m. Due to wind, a towel from a terrace on a nearby building fell into the garden. Then find the probability of the event that it fell in the lake.

Solution:
Area of the rectangular garden
= length × breadth
= 77 × 50
∴ Area of the rectangular garden = 3850 sq.m
Radius of the lake = \(\frac { 14 }{ 2 } \) = 7 m

∴ The probability of the event that the towel tell in the lake is \(\frac { 1 }{ 25 } \).

Question 10.
In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at
i. 8.
ii. an odd number.
iii. a number greater than 2.
iv. a number less than 9.

Solution:
Sample space (S) = {1,2, 3, 4, 5, 6, 7, 8}
∴ n(S) = 8
i. Let A be the event that the spinning arrow comes to rest at 8.

ii. Let B be the event that the spinning arrow comes to rest at an odd number.

iii. Let C be the event that the spinning arrow comes to rest at a number greater than 2.

iv. Let D be the event that the spinning arrow comes to rest at a number less than 9.
∴ D = {1,2, 3, 4, 5, 6, 7, 8}

Question 11.
There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
i. a natural number.
ii. a number less than 1.
iii. a whole number.
iv. a number greater than 5.
Solution:
Sample space (S) = {0, 1, 2, 3, 4, 5}
∴ n(S) = 6

i. Let A be the event that the card drawn shows a natural number.
∴ A = {1,2,3,4,5}
∴ n(A) = 5

ii. Let B be the event that the card drawn shows a number less than 1.
∴ B = {0}
∴ n(B) = 1

iii. Let C be the event that the card drawn shows a whole number.
∴ C = {0,1, 2, 3, 4, 5}
∴ n(C) = 6

iv. Let D be the event that the card drawn shows a number greater than 5.
Here, the greatest number is 5.
∴ Event D is an impossible event.
∴ D = { }
∴ n(D) = 0

Question 12.
A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is:
i. red.
ii. not red.
iii. either red or white.
Solution:
Let the three red balls be R₁, R₂, R₃, three white balls be W₁, W₂, W₃ and three green balls be G₁, G₂, G₃.
∴ Sample space,
S = {R₁, R₂, R₃, W₁, W₂, W₃, G₁, G₂, G₃}
∴ n(S) = 9

i. Let A be the event that the ball drawn is red.
∴ A = {R₁, R₂, R₃}
∴ n(A) = 3

ii. Let B be the event that the ball drawn is not red.
B = {W₁,W₂,W₃,G₁,G₂,G₃}
∴ n(B) = 6

iii. Let C be the event that the ball drawn is red or white.
∴ C = {R₁, R₂, R₃, W₁, W₂, W₃}
∴ n(C) = 6

Question 13.
Each card bears one letter from the word ‘mathematics’. The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.
Solution:
Sample space
= {m, a, t, h, e, m, a, t, i, c, s}
∴ n(S) = 11
Let A be the event that the card drawn bears the letter ‘m’
∴ A = {m, m}
∴ n(A) = 2

∴ The probability that a card drawn bears letter ‘m’ is \(\frac { 2 }{ 11 } \).

Question 14.
Out of 200 students from a school, 135 like Kabaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected dosen’t like Kabaddi.
Solution:
Total number of students in the school = 200
∴ n(S) = 200
Number of students who like Kabaddi = 135
∴ Number of students who do not like Kabaddi
= 200 – 135 = 65
Let A be the event that the student selected does not like Kabaddi.
∴ n(A) = 65

∴ The probability that the student selected doesn’t like kabaddi is \(\frac { 13 }{ 40 } \).

Question 15.
A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a:
i. prime number.
ii. multiple of 4.
iii multiple of 11.
Solution:
Sample space
(S) = {10, 11, 12, 13, 14,
20, 21, 22, 23, 24,
30, 31, 32, 33, 34,
40, 41, 42, 43, 44}
∴ n(S) = 20

i. Let A be the event that the number so formed is a prime number.
∴ A = {11,13,23,31,41,43}
∴ n(A) = 6

ii. Let B be the event that the number so formed is a multiple of 4.
∴ B = {12,20,24,32,40,44}
∴ n(B) = 6

iii. Let C be the event that the number so formed is a multiple of 11.
∴ C = {11,22,33,44}
∴ n(C) = 4

Question 16.
The faces of a die bear numbers 0,1, 2, 3,4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.
Solution:
Sample space,
S = {(0, 0), (0,1), (0,2),
(1,0), (1,1), (1,2),
(2,0), (2,1), (2,2),
(3.0), (3,1), (3,2),
(4.0), (4,1), (4,2),
(5.0), (5,1), (5,2),
∴ n(S) = 36
Let A be the event that the product of digits on the upper face is zero.
∴ A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1,0), (2, 0), (3,0), (4, 0), (5,0)}
∴ n(A) = 11

∴ The probability that the product of the digits on the upper face is zero is \(\frac { 11 }{ 36 } \).

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.1 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.1 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.

Solution:

∴ The mean of the time spent by the students for their studies is 4.36 hours.

Question 2.
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by ‘assumed mean’ method.

Solution:
Let us take the assumed mean (A) = 550

∴ The mean of the toll paid by the drivers is ₹ 521.43.

Question 3.
A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.

Solution:

∴ The mean of the milk sold is 2.82 litres.

Question 4.
A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by ‘assumed mean’ method.

Solution:
Let us take the assumed mean (A) = 37.5

∴ The mean of the production of oranges is ₹ 35310.

Question 5.
A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of he funds by ‘step deviation’ method.

Solution:
Here, we take A = 1250 and g = 500

∴ The mean of the funds collected is ₹ 987.5.

Question 6.
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by ‘step deviation’ method.

Solution:
Here, we take A = 2500 and g = 1000.

∴ The mean of the weekly wages is ₹ 3070.

Question 1.
The daily sale of 100 vegetable vendors is given in the following table. Find the mean of the sale by direct method. (Textbook pg. no. 133 and 134)

Solution:

The mean of the sale is 2150.

Question 2.
The amount invested in health insurance by 100 families is given in the following frequency table. Find the mean of investments using direct method and assumed mean method. Check whether the mean found by the two methods is the same as calculated by step deviation method (Ans: ₹ 2140). (Textbook pg. no. 135 and 136)

Solution:

∴ The mean of investments in health insurance is ₹ 2140.
Assumed mean method:

∴ The mean of investments in health insurance is ₹ 2140.
∴ Mean found by direct method and assumed mean method is the same as calculated by step deviation method.

Question 3.
The following table shows the funds collected by 50 students for flood affected people. Find the mean of the funds.

If the number of scores in two consecutive classes is very low, it is convenient to club them. So, in the above example, we club the classes 0 – 500, 500 – 1000 and 2000 – 2500, 2500 – 3000. Now the new table is as follows

i. Solve by direct method.
ii. Verily that the mean calculated by assumed mean method is the same.
iii. Find the mean in the above example by taking A = 1750. (Textbook pg. no. 137)
Solution:
i. Direct method:

∴ The mean of the funds is ₹ 1390.

ii. Assumed mean method:
Here, A = 1250

∴ The mean calculated by assumed mean method is the same.

iii. Step deviation method:
Here, we take A = 1750 and g = 250

∴ The mean of the funds is ₹ 1390.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

10th Standard Maths 1 Practice Set 5.4 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 5.4 Chapter 5 Probability Questions With Answers Maharashtra Board

Question 1.
If two coins are tossed, find the probability of the following events.
i. Getting at least one head.
ii. Getting no head.
Solution:
Sample space,
S = {HH, HT, TH, TT}
∴ n(S) = 4

i. Let A be the event of getting at least one head.
∴ A = {HT, TH, HH}
∴ n(A) = 3
∴ P(A) = \(\frac { n(A) }{ n(S) } \)
∴ P(A) = \(\frac { 3 }{ 4 } \)

ii. Let B be the event of getting no head.
∴ B = {TT}
∴ n(B) = 1
∴ P(B) = \(\frac { n(B) }{ n(S) } \)
∴ P(B) = \(\frac { 1 }{ 4 } \)
∴ P(A) = \(\frac { 3 }{ 4 } \); P(B) = \(\frac { 1 }{ 4 } \)

Question 2.
If two dice are rolled simultaneously, find the probability of the following events.
i. The sum of the digits on the upper faces is at least 10.
ii. The sum of the digits on the upper faces is 33.
iii. The digit on the first die is greater than the digit on second die.
Solution:
Sample space,
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

i. Let A be the event that the sum of the digits on the upper faces is at least 10.
∴ A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 6
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 6 }{ 36 } \)
∴ P(A) = \(\frac { 1 }{ 6 } \)

ii. Let B be the event that the sum of the digits on the upper faces is 33.
The sum of the digits on the upper faces can be maximum 12.
∴ Event B is an impossible event.
∴ B = { }
∴ n(B) = 0
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 0 }{ 36 } \)
∴ P(B) = 0

iii. Let C be the event that the digit on the first die is greater than the digit on the second die.
C = {(2, 1), (3, 1), (3,2), (4,1), (4,2), (4, 3), (5, 1), (5,2), (5,3), (5,4), (6,1), (6,2), (6, 3), (6, 4), (6, 5),
∴ n(C) = 15
∴ P(C) = \(\frac { n(c) }{ n(S) } \) = \(\frac { 15 }{ 36 } \)
∴ P(C) = \(\frac { 5 }{ 12 } \)
∴ P(A) = \(\frac { 1 }{ 6 } \) ; P(B) = 0; P(C) = \(\frac { 5 }{ 12 } \)

Question 3.
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn:
i. shows an even number.
ii. shows a number which is a multiple of 5.
Solution:
Sample space,
S = {1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15}
∴ n(S) = 15

i. Let A be the event that the ticket drawn shows an even number.
∴ A = {2, 4, 6, 8, 10, 12, 14}
∴ n(A) = 7
∴ P(A) = \(\frac { n(A) }{ n(S) } \)
∴ P(A) = \(\frac { 7 }{ 15 } \)

ii. Let B be the event that the ticket drawn shows a number which is a multiple of 5.
∴ B = {5, 10, 15}
∴ n(B) = 3
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 3 }{ 15 } \)
∴ P(B) = \(\frac { 1 }{ 5 } \)
∴ P(A) = \(\frac { 7 }{ 15 } \) ; P(B) = \(\frac { 1 }{ 5 } \)

Question 4.
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
i. an odd number?
ii. a multiple of 5?
Solution:
Sample space
(S) = {23, 25, 27, 29,
32, 35, 37, 39,
52, 53, 57, 59,
72, 73, 75, 79,
92, 93, 95, 97}
∴ n(S) = 20
i. Let A be the event that the number formed is an odd number.
∴ A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75,79,93,95,97}
∴ n(A) = 16
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 16 }{ 20 } \)
∴ P(A) = \(\frac { 4 }{ 5 } \)

ii. Let B be the event that the number formed is a multiple of 5.
∴ B = {25,35,75,95}
∴ n(B) = 4
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 4 }{ 20 } \)
∴ P(B) = \(\frac { 1 }{ 5 } \)
∴ P(A) = \(\frac { 4 }{ 5 } \) ; P(B) = \(\frac { 1 }{ 5 } \)

Question 5.
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is
i. an ace.
ii. a spade.
Solution:
There are 52 playing cards.
∴ n(S) = 52
i. Let A be the event that the card drawn is an ace.
∴ n(A) = 4
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 4 }{ 52 } \)
∴ P(A) = \(\frac { 1 }{ 13 } \)

ii. Let B be the event that the card drawn is a spade.
∴ n(B) = 13
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 13 }{ 52 } \)
∴ P(B) = \(\frac { 1 }{ 4 } \)
∴ P(A) = \(\frac { 1 }{ 13 } \) ; P(B) = \(\frac { 1 }{ 4 } \)

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

Maharashtra State Board 10th Std Sanskrit Textbook Solutions

Maharashtra State Board Class 10th SSC Sanskrit Aamod Digest Std 10 Pdf Free Download and Anand Sanskrit Composite Digest Pdf Std 10.

Maharashtra State Board 10th Std Sanskrit Aamod Textbook Solutions

Maharashtra State Board Aamod Sanskrit Book 10th Class Pdf Download, Sanskrit Aamod Class 10 Textbook Solutions PDF Free Download.

Sanskrit Aamod Class 10 Textbook Solutions

• Chapter 1 आद्यकृषक: पृथुवैन्यः (गद्यम्)
• Chapter 2 व्यसने मित्रपरीक्षा (गद्यम्)
• Chapter 3 सूक्तिसुधा (पद्यम्) (कण्ठस्थीकरणार्थम्)
• Chapter 4 अमूल्यं कमलम् (गद्यम्)
• Chapter 5 स एव परमाणुः (संवादः)
• Chapter 6 युग्ममाला (पद्यम्) (कण्ठस्थीकरणार्थम्)
• Chapter 7 संस्कृतनाट्ययुग्मम् (संवादः) (सरलार्थलेखनार्थम्)
• Chapter 8 वाचनप्रशंसा (पद्यम्) (सरलार्थलेखनार्थम्)
• Chapter 9 धेनोर्व्याघ्रः पलायते (गद्यम्)
• Chapter 10 नदीसूक्तम् (संवादः) (सरलार्थलेखनार्थम्)
• Chapter 11 जटायुशौर्यम् (पद्यम्)
• Chapter 12 आदिशङ्कराचार्यः (गद्यम्)
• Chapter 13 चित्रकाव्यम् (पद्यम्) (कण्ठस्थीकरणार्थम्)
• Chapter 14 प्रतिपदं संस्कृतम् (संवादः)
• Chapter 15 मानवताधर्मः (पद्यम्) (सरलार्थलेखनार्थम्)

Maharashtra State Board 10th Std Sanskrit Anand Textbook Solutions

Maharashtra State Board 10th Standard Sanskrit Anand Digest Pdf, 10th Composite Sanskrit Digest Pdf download.

Anand Sanskrit Composite Digest Pdf Std 10

• Chapter 1 आद्यकृषक: पृथुवैन्यः (गद्यम्)
• Chapter 2 व्यसने मित्रपरीक्षा (गद्यम्)
• Chapter 3 सूक्तिसुधा (पद्यम्) (कण्ठस्थीकरणार्थम्)
• Chapter 4 स एव परमाणुः (संवादः)
• Chapter 5 युग्ममाला (पद्यम्) (कण्ठस्थीकरणार्थम्)
• Chapter 6 संस्कृतनाट्यस्तबकः (संवादः) (सरलार्थलेखनार्थम्)
• Chapter 7 वाचनप्रशंसा (पद्यम्) (सरलार्थलेखनार्थम्)
• Chapter 8 नदीसूक्तम् (संवादः) (सरलार्थलेखनार्थम्)
• Chapter 9 आदिशङ्कराचार्यः (गद्यम्)
• Chapter 10 चित्रकाव्यम् (पद्यम्) (कण्ठस्थीकरणार्थम्)
• Chapter 11 मानवताधर्मः (पद्यम्) (सरलार्थलेखनार्थम्)

Maharashtra State Board Class 10 Textbook Solutions