Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 5- Probability

Class 10: Maths Chapter 5 solutions. Complete Class 10 Maths Chapter 5 Notes.

Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 5- Probability

Maharashtra Board 10th Maths Chapter 5, Class 10 Maths Chapter 5 solutions

Practice Set 5.1

Question 1.
How many possibilities are there in each of the following?
i. Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
ii. Any day of a week is to be selected randomly.
iii. Select one card from the pack of 52 cards.
iv. One number from 10 to 20 is written on each card. Select one card randomly.
Solution:
i. Here, 8 sites of Maharashtra are given.
∴ There are 8 possibilities in a random experiment of visiting a site out of 8 sites in Maharashtra.

ii. There are 7 days in a week.
∴ There are 7 possibilities in a random experiment of selecting a day of the week.

iii. There are 52 cards in a pack of cards.
∴ There are 52 possibilities in a random experiment of selecting one card from the pack of 52 cards.

iv. There are 11 cards numbered from 10 to 20.
∴ There are 11 possibilities in a random experiment of selecting one card from the given set of cards.

Question 1.
In which of the following experiments possibility of expected outcome is more? (Textbook pg, no. 116)
i. Getting 1 on the upper face when a die is thrown.
ii. Getting head by tossing a coin.
Solution:
i. On a die there are 6 numbers.
∴ There are 6 possibilities of getting any one number from 1 to 6 on the upper face i.e. 16 is the possibility.

ii. There are two possibilities (H or T) on tossing a coin i.e. 12 possibility.
∴ In the second experiment, the possibility of expected outcome is more.

Question 2.
Throw a die, once. What are the different possibilities of getting dots on the upper face? (Textbook pg. no. 114)
Answer:
There are six different possibilities of getting dots on the upper face. They are

Practice Set 5.2

Question 1.
For each of the following experiments write sample space ‘S’ and number of sample Point n(S)
i. One coin and one die are thrown simultaneously.
ii. Two digit numbers are formed using digits 2,3 and 5 without repeating a digit.
Solution:
i. Sample space,
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)}
∴ n(S) =12
ii. Sample space,
S = {23,25,32, 35, 52, 53}
∴ n(S) = 6

Question 2.
The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.

Solution:
There are total six colours on the disc.
Sample space,
S = {Red, Orange, Yellow, Blue, Green, Purple}
∴ n(S) = 6
∴ Arrow may stop on any one of the six colours.

Question 3.
In the month of March 2019, find the days on which the date is a multiple of 5. (see the given page of the calendar).

Solution:
Dates which are multiple of 5:
5,10, 15,20,25,30
∴ S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
∴ The days on which the date will be a multiple of 5 are Tuesday, Sunday, Friday, Wednesday, Monday and Saturday.

Question 4.
Form a ‘Road safety committee’ of two, from 2 boys (B₁ B₂) and 2 girls (G₁, G₂). Complete the following activity to write the sample space.
Solution:

Question 1.
Sample Space

• The set of all possible outcomes of a random experiment is called sample space.
• It is denoted by ‘S’ or ‘Ω’ (omega).
• Each element of a sample space is called a sample point.
• The number of elements in the set S is denoted by n(S).
• If n(S) is finite, then the sample space is called a finite sample space.

Some examples of finite sample space. (Textbook pg. no, 117)
Solution:
 

Practice Set 5.3

i. One die is rolled,
Event A: Even number on the upper face.
Event B: Odd number on the upper face.
Event C: Prime number on the upper face.

ii. Two dice are rolled simultaneously,
Event A: The sum of the digits on upper faces is a multiple of 6.
Event B: The sum of the digits on the upper faces is minimum 10.
Event C: The same digit on both the upper faces.

iii. Three coins are tossed simultaneously.
Condition for event A: To get at least two heads.
Condition for event B: To get no head.
Condition for event C: To get head on the second coin.

iv. Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A: The number formed is even.
Condition for event B: The number is divisible by 3.
Condition for event C: The number formed is greater than 50.

v. From three men and two women, environment committee of two persons is to be formed.
Condition for event A: There must be at least one woman member.
Condition for event B: One man, one woman committee to be formed.
Condition for event C: There should not be a woman member.

vi. One coin and one die are thrown simultaneously.
Condition for event A: To get head and an odd number.
Condition for event B: To get a head or tail and an even number.
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
Solution:
i. Sample space (S) = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Condition for event A: Even number on the upper face.
∴ A = {2,4,6}
∴ n(A) = 3
Condition for event B: Odd number on the upper face.
∴ B = {1, 3, 5}
∴ n(B) = 3
Condition for event C: Prime number on the upper face.
∴ C = {2, 3, 5}
∴ n(C) = 3

ii. 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
Condition for event A: The sum of the digits on the upper faces is a multiple of 6.
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
∴ n(A) = 6

Condition for event B: The sum of the digits on the upper faces is minimum 10.
B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 6

Condition for event C: The same digit on both the upper faces.
C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(C) = 6

iii. Sample space,
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ n(S) = 8

Condition for event A: To get at least two heads.
∴ A = {HHT, HTH, THH, HHH}
∴ n(A) = 4

Condition for event B: To get no head.
∴ B = {TTT}
∴ n(B) = 1

Condition for event C: To get head on the second coin.
∴ C = {HHH, HHT, THH, THT}
∴ n(C) = 4

iv. Sample space (S) = {10, 12, 13, 14, 15,
20, 21, 23, 24, 25,
30, 31, 32, 34, 35,
40, 41, 42, 43,
45, 50, 51, 52, 53, 54}
∴ n(S) = 25
Condition for event A: The number formed is even
∴ A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54)
∴ n(A) = 13
Condition for event B: The number formed is divisible by 3.
∴ B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
∴ n(B) = 9
Condition for event C: The number formed is greater than 50.
∴ C = {51,52, 53,54}
∴ n(C) = 4

v. Let the three men be M₁, M₂, M₃ and the two women be W₁, W₂.
Out of these men and women, a environment committee of two persons is to be formed.
∴ Sample space,
S = {M₁M₂, M₁M₃, M₁W₁, M₁W₂, M₂M₃, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}
∴ n(S) = 10
Condition for event A: There must be at least one woman member.
∴ A = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}
∴ n(A) = 7
Condition for event B: One man, one woman committee to be formed.
∴ B = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₂, M₃W₂}
∴ n(B) = 6
Condition for event C: There should not be a woman member.
∴ C = {M₁M₂, M₁M₃, M₂M₃}
∴ n(C) = 3

vi. Sample space,
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)}
∴ n(S) = 12
Condition for event A: To get head and an odd number.
∴ A = {(H, 1), (H, 3), (H, 5)}
∴ n(A) = 3
Condition for event B: To get a head or tail and an even number.
∴ B = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6)}
∴ n(B) = 6
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
The greatest number on the upper face of a die is 6.
∴ Event C is an impossible event.
∴ C = { }
∴ n(C) = 0

Practice Set 5.4

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) = n(A)n(S)
∴ P(A) = 34

ii. Let B be the event of getting no head.
∴ B = {TT}
∴ n(B) = 1
∴ P(B) = n(B)n(S)
∴ P(B) = 14
∴ P(A) = 34; P(B) = 14

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) = n(A)n(S) = 636
∴ P(A) = 16

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) = n(B)n(S) = 036
∴ 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) = n(c)n(S) = 1536
∴ P(C) = 512
∴ P(A) = 16 ; P(B) = 0; P(C) = 512

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) = n(A)n(S)
∴ P(A) = 715

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) = n(B)n(S) = 315
∴ P(B) = 15
∴ P(A) = 715 ; P(B) = 15

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) = n(A)n(S) = 1620
∴ P(A) = 45

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) = n(B)n(S) = 420
∴ P(B) = 15
∴ P(A) = 45 ; P(B) = 15

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) = n(A)n(S) = 452
∴ P(A) = 113

ii. Let B be the event that the card drawn is a spade.
∴ n(B) = 13
∴ P(B) = n(B)n(S) = 1352
∴ P(B) = 14
∴ P(A) = 113 ; P(B) = 14

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Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 5- Probability

Download PDF: Maharashtra Board Solutions for Class 10-Maths (Part 1): Chapter 5- Probability PDF

Chapterwise Maharashtra Board Solutions Class 10 Maths (Part 1) :

• Chapter 1- Linear Equations in Two Variables
• Chapter 2- Quadratic Equations
• Chapter 3- Arithmetic Progression
• Chapter 4- Financial Planning
• Chapter 5- Probability
• Chapter 6- Statistics

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