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Learn to solve these important probability questions with answers:-

Probability is an important topic of the aptitude section in all competitive exams. There are always 1-2 questions either from this chapter or the related chapter of permutation and combinations. Without good understanding of permutations and combinations, solving the probability questions correctly is unimaginable. So I hope that you have already completed that chapter.

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

1. 3/4
2. 1/4
3. 3/8
4. 7/8

When 3 unbiased coins are tossed then there total 8 outcomes as {HTT, HTH, HHT, HHH, TTT, THT, THH, TTH}. Out of this set there is only one case where all coins are head. In other 7 outcomes there are either no head, 1 head or 2 heads. So our answer is favorable number of outcomes divided by total number of outcomes = 7/8.

What is the probability of getting a sum 9 from two throws of a dice?

1. 1/6
2. 1/8
3. 1/9
4. 1/12

The total number of outcomes in two throws of a dice is 6*6=36. And the favorable events in this case sum to be 9 will be calculated as follows:-

9 = 6 + 3 (6 in first throw and 3 in the second throw)

9 = 3 + 6

9 = 5 + 4

9 = 4 + 5

So there will be total 4 favorable outcomes. Hence the required probability = 4/36 = 1/9. Please provide your feedback below if you like the explanations about these probability questions and answers.

Two dice are tossed. The probability that the total score is a prime number is:

1. 1/6
2. 5/12
3. 1/2
4. 7/9

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

1. 1/10
2. 2/5
3. 2/7
4. 5/7

Obviously there are 10 favorable outcomes as number of prizes are 10. And total number of outcomes is 10+25 = 35.

Hence the required probability is 10/35 = 2/7.

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

1. 1/15
2. 25/57
3. 35/256
4. 1/221

Here the favorable outcomes can be calculated as selecting 2 cards out of 4 king cards. That can be done in ⁴C2 ways which comes to be (4*3)/2 = 6 ways.

Total number of outcomes, means selecting 2 cards out of 52 cards = 52C2 ways. That is (52*51)/2 = 26*51

Hence required probability is = 6 / 26*51 = 1/221.

Dear friends, if you have any questions or feedback regarding probability questions and answers then please don’t hesitate commenting me down on this page and I will surely get back as soon as possible.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is:

1. 21/46
2. 25/117
3. 1/50
4. 3/25

There are total 25 students in class and we have to select 3 students. Total number of ways to do so is 25C3=(25*24*23)/(3*2*1) = 2300

Now let’s see how many are the favorable ways as per the question. As we have to select 1 girl out of 10 girls AND 2 boys out of 15 boys so this can be done in 10C1* 15C2 = 10*15*14/2 = 1050.

The required probability here is  = 1050/2300 = 21/46.

Please note that we have used multiplication for AND operator.

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

1. 10/21
2. 11/21
3. 2/7
4. 5/7

This probability questions asking us not to select any blue balls that means we can only select either red or green balls.

Our favorable outcomes are the number of ways to select 2 balls out of 5 balls (2 red and 3 green). = 5C2  = 10.

And total number ways to select 2 balls out of total 7 balls = 7C2 = 7*6/2 = 21

Hence the required probability is = 10/21

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and other is a heart, is:

1. 3/20
2. 29/34
3. 47/100
4. 13/102

In 52 cards there are 13 spades and 13 hearts. We have to select 1 spade and 1 heart out of 13 cards each to find out our favorable outcomes.  That can be done as 13C1 * 13C1 =13*13.

And total number of ways to select 2 cards out of 52 cards is 52C2 = 52*51/2 = 26*51

Hence the required probability  = (13*13)/(26*51) = 13/102.

A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

1. 3/4
2. 4/7
3. 1/8
4. 3/7

Here in this very simple probability question, we have number of ways to select 1 white ball out of 8 white balls is = 8

And total number of ways to select 1 ball out of total 14 balls is = 14.

Hence the required probability is 8/14 = 4/7.

A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that they are of the same color ? (Also find the probability that they are not of the same color ?)

Here for both balls to be of the same color, they can either be white or both of them can be black. So 6C2 + 4C2 will be the favorable outcomes that is = 15+6 = 21

And total number of ways to select 2 balls out of 10 total balls = 10C2 = 45.

Hence the probability of both balls to be of the same color = 21/45 =7/15.

And further if we have to find the probability that both balls are of different colors then we can subtract above result from 1 as there can be only 2 cases i.e. either both balls are of different colors or they are of the same color. Hence probability in this case will be 1 – 7/15 = 8/15.