RBSE Maths Class 10 Chapter 18: Probability Important Questions and Solutions

RBSE Class 10 Maths Chapter 18 – Probability Important questions and solutions are provided here. All these important questions and solutions are given here with detailed explanations. The RBSE Class 10 important questions and solutions available at BYJU’S cover all the concepts, which are important from the examination point of view.

Chapter 18 of RBSE Class 10 contains only one exercise which involves questions on finding the probability of various events. In this chapter, data of different experiments have been given for which the probability for related events is to be calculated. All these experiments given here are realistic in nature, that means they can be observed in our daily life in various situations.

RBSE Maths Chapter 18: Exercise 18.1 Textbook Important Questions and Solutions

Question 1: In a throw of a die, determine the probability of getting a number more than four.

Solution:

When a die is thrown, the possible outcomes = {1, 2, 3, 4, 5, 6}

Total number of outcomes = 6

Outcomes greater than 4 = {5, 6}

Number of favourable outcomes = 2

Probability of getting a number greater than 4 = 2/6 = 1/3

Question 2: A coin is tossed twice. Find the probability of getting tails both times.

Solution:

Given,

Coins are tossed twice.

Possible outcomes = {HH, HT, TH, TT}

Total number of outcomes = 4

Number of favourable outcomes for getting heads both the times = 1

i.e. HH

Therefore, the required probability = 1/4

Question 3: One number is selected at random from natural numbers from 1 to 17. Find the probability that the number is prime.

Solution:

Total number of possible outcomes = 17

i.e. natural numbers from 1 to 17

Prime numbers from 1 to 17 = {2, 3, 5, 7, 11, 13, 17}

Number of favourable outcomes = 7

Therefore, the required probability = 7/17

Question 4: Find the probability of throwing head or tail alternatively in 3 successive tossing of a coin.

Solution:

Given that, a coin is tossed three times.

Total number of possible outcomes = 8

i.e. {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Outcomes in which heads occur alternatively are: HTH, THT

Thus, the number of favourable outcomes = 2

Therefore, the required probability = 2/8 = 1/4

Question 5: Find the probability that a non-leap year should have only 52 Sundays.

Solution:

We know that, number of days in a non-leap year = 365

365 days = 52 weeks + 1 odd day

Therefore, a non-leap year will have 52 Sundays.

1 odd day may be one of the following 7 days:

Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday

Thus, total number of possible outcomes = 7

Number of favourable outcomes for which 1 odd day is not a Sunday = 6

P(non-leap year has 52 Sundays) = 6/7

Question 6: If P(A) = 0.65, then what will be the probability of “not A”?

Solution:

Given,

P(A) = 0.65

We know that,

P(A) + P(not A) = 1

0.65 + P(not A) = 1

P(not A) = 1 – 0.65

P(not A) = 0.35

Therefore, the probability of “not A” is 0.35.

Question 7: On tossing two coins, find the probability of getting at most one tail.

Solution:

Given that, two coins are tossed.

Total number of possible outcomes = 4

i.e. {HH, HT, TH, TT}

Let A be the event of getting at most one tail.

Number of outcomes favourable to A = 3

i.e. {HT, TH, TT}

P(A) = ¾

Hence, the required probability is ¾.

Question 8: A die is thrown twice. What will be the probability that the sum on the faces is:

(i) 9 (ii) 13

Solution:

Given that, a die is thrown twice.

Total number of outcomes = 62 = 36

(i) Let A be the event of getting a sum 9.

Number of favourable outcomes to A = 4

i.e. {(3, 6), (6, 3), (4, 5), (5, 4)}

P(A) = 4/36 = 1/9

Therefore, the probability of getting the sum on the faces of dice = 1/9

(ii) Let B be the event of getting a sum 13.

We know that, the maximum possible sum of numbers on dice = 12 (i.e. 6 + 6)

Thus, the sum of numbers on dice is not possible.

Hence, the probability of getting the sum of the faces of dices = 0

Question 9: A bag contains 5 red and 3 white balls. From this bag, one ball is drawn randomly. What will be the probability that the drawn ball is

(i) White (ii) Not white

Solution:

Given that, a bag contains 5 red and 3 white balls.

Total number of possible outcomes = 8

i.e. 5 + 3 = 8

(i) Let E be the event of drawing a white ball from the bag.

Number of outcomes favourable to E = 3

P(E) = ⅜

(ii) Let the event of not drawing a ball which is not a white ball be “not E”.

P(not E) = 1 – P(E)

= 1 – ⅜

= (8 – 3)/8

= ⅝

Therefore, the probability of drawing a white ball is ⅜ and the probability of not drawing a white ball is ⅝.

Question 10: A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting the following.

(i) Jack of red colour

(ii) Card of red colour

(iii) Ace of heart

(iv) Card of spade

Solution:

Total number of possible outcomes = 52

(i) Total number of jack of red colour cards = 2

P (getting a jack of red colour card) = 2/52 = 1/26

(ii) Total number of red colour cards = 26

P (getting a card of red colour) = 26/52 = 1/2

(iii) Total number of ace of heart cards = 1

P (getting an ace of hearts) = 1/52

(iv) Total number of cards of spades = 13

P (getting a card of spade) = 13/52 = 1/4

RBSE Maths Chapter 18: Additional Important Questions and Solutions

Question 1: A bag contains 15 cards. The numbers 1, 2, 3, 4,…, 15 are printed on them. A card is drawn randomly from the bag. Find the probability that the number on the card is

(i) a prime number

(ii) a number is divisible by 2

Solution:

Given that, a bag contains 15 cards with numbers from 1 to 15 printed on them.

Total number of possible outcomes = 15

(i) Let A be the event of drawing a prime numbered card.

Number of outcomes favourable to A = 6

i.e. {2, 3, 5, 7, 11, 13}

P (drawing a card with prime number) = 6/15

(ii) Let B be the event of drawing a card with the number divisible by 2.

Number of outcomes favourable to B = 7

i.e. {2, 4, 6, 8, 10, 12, 14}

P (drawing a card with a number divisible by 2) = 7/15

Question 2: In a bag, one white ball, two black balls and three red balls of the same size are placed. A ball is drawn at random from this bag. Find the probability of the following.

(i) Ball is white

(ii) Ball is not black

(iii) Ball is red

Solution:

Given that, a bag contains 1 white, 2 black and 3 red balls.

Total number of possible outcomes = 6

i.e. 1 + 2 + 3

(i) Total number of white balls = 1

P(getting a white ball) = ⅙

(ii) Total number of balls other than black = 1 + 3 = 4

i.e. 1 white and 3 red

P(getting a ball which is not black) = 4/6 = ⅔

(iii) Total number of red balls = 3

P(getting a red ball) = 3/6 = 1/2

Question 3: If probability of “not E” = 0.95, then find P(E).

Solution:

Given,

P(not E) = 0.95

We know that,

P(E) + P(not E) = 1

P(E) + 0.95 = 1

P(E) = 1 – 0.95

P(E) = 0.05

Question 4: A box contains 7 red marbles, 10 white marbles and 5 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

(i) not red?

(ii) white?

(iii) green?

Solution:

Total number of possible outcomes = 22

i.e. 7 red + 10 white + 5 green marbles = 22 marbles

(i) Total number of marbles other than red = 10 + 5 = 15

i.e. 10 white and 5 green marbles

P(drawing a marble which is not red) = 15/22

(ii) Total number of white marbles = 10

P(drawing a white marble) = 10/22 = 5/11

(iii) Total number of green marbles = 5

P(drawing a green marble) = 5/22

Question 5: A piggy bank contains hundred coins of Rs. 1, twenty five coins of Rs. 2, fifteen coins of Rs. 5 and ten coins of Rs. 10. If it is likely that one coin will fall, when the bank is turned upside down, what is the probability that the coin

(i) will be a Rs, 2 coin?

(ii) will not be a Rs. 5 coin?

Solution:

Given,

A piggy bank contains hundred coins of Rs. 1, twenty five coins of Rs. 2, fifteen coins of Rs. 5 and ten coins of Rs. 10.

Total number of possible outcomes = 150

i.e. 100 + 25 + 15 + 10 = 50 coins

(i) Total number of Rs. 2 coins = 25

P(getting a Rs. 2 coin) = 25/150 = ⅙

(ii) Total number of coins other than Rs. 5 coin = 150 – 15 = 135

P(getting a coin which is not a Rs. 5 coin) = 135/150 = 9/10

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