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QUESTION: 1

A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Solution:

QUESTION: 2

A die is rolled twice. What is the probability of getting a sum equal to 9?

Solution:

QUESTION: 3

Three coins are tossed. What is the probability of getting at most two tails?

Solution:

QUESTION: 4

When tossing two coins once, what is the probability of heads on both the coins?

Solution:

QUESTION: 5

What is the probability of getting a number less than 4 when a die is rolled?

Solution:

QUESTION: 6

A bag contains 4 black, 5 yellow and 6 green balls. Three balls are drawn at random from the bag. What is the probability that all of them are yellow?

Solution:

QUESTION: 7

One card is randomly drawn from a pack of 52 cards. What is the probability that the card drawn is a face card(Jack, Queen or King)

Solution:

Total number of cards, n(S) = 52

Total number of face cards, n(E) = 12

QUESTION: 8

A dice is thrown. What is the probability that the number shown in the dice is divisible by 3?

Solution:

Total number of outcomes possible when a die is rolled, n(S) = 6 (? 1 or 2 or 3 or 4 or 5 or 6)

E = Event that the number shown in the dice is divisible by 3 = {3, 6}

Hence, n(E) = 2

QUESTION: 9

John draws a card from a pack of cards. What is the probability that the card drawn is a card of black suit?

Solution:

Total number of cards, n(S) = 52

Total number of black cards, n(E) = 26

QUESTION: 10

There are 15 boys and 10 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?

Solution:

Let S be the sample space.

n(S) = Total number of ways of selecting 3 students from 25 students = ^{25}C_{3}

Let E = Event of selecting 1 girl and 2 boys

n(E) = Number of ways of selecting 1 girl and 2 boys

15 boys and 10 girls are there in a class.

We need to select 2 boys from 15 boys and 1 girl from 10 girls

Number of ways in which this can be done = ^{15}C_{2} × ^{10}C_{1}

Hence n(E) = ^{15}C_{2} × ^{10}C_{1}

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