Introduction
Probability is the ratio of expected outcomes to all possible results to determine the likelihood that an action will occur. It's one of the most obvious uses of arithmetic because it can assist people in making predictions, and it's frequently used in statistics.
Predicting whether a tossed coin will land on heads or tails is a simple example of probability maths. The probability formula is all possible outcomes/number of desired outcomes.
There is only one desirable outcome (for example, heads) out of two possible outcomes for coin tossing. As a result, the chance of a cointoss coming up heads is 1/2.
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Let us now go through some of the crucial gate questions on probability.
Gate Questions on Probability
Below are some commonly asked gate questions on probability in the GATE examination.
1. A fairsided dice is rolled three times. The likelihood of an exact one odd number appearing in one of the three outcomes is
 1/6
 3/8
 1/8

1/2
Ans. b) 3/8
Explanation: The question is a binomial experiment with two outcomes: Number is Even(E) or Number is Odd(O).
P(E) = P(O) = 1/2
P(exactly one odd number) = ^{3}C_{1} * P(E)^{2} * P(O)^{1} = â…ś
2. There's a 0.5 chance that it'll rain today. Tomorrow has a 0.6 percent chance of raining. Today or tomorrow, there is a 0.7 percent chance of rain. What are the chances of it raining today and tomorrow?
 0.3
 0.25
 0.35

0.4
Ans. d) 0.4
Explanation: Let TD represent today's rain probability and TM represent tomorrow's.
Probability (TD intersection TM) = probability ( TD ) + probability (TM)  Probability (TD union TM).
Probability (TD intersection TM) = 0.5 + 0.6  0.7 = 0.4
3. The keys 44, 45, 79, 55, 91, 18, and 63 are inserted into a table indexed from 0 to 6 using a hash function h defined as h(key)=key mod 7 and linear probing. What will the placement of key number 18 be?
 3
 4
 5

6
Ans. c) 5
Explanation:
h(44) = 44mod7 = 2
h(45) = 45mod7 = 3
h(79) = 79mod7 = 2, but 2 is already filled by 44, therefore linear probing is used. As a result, the number 79 will be assigned 4.
h(55) = 55mod7 = 6
h(91) = 91mod7 = 0
h(18) = 18mod7 = 4, but 4 is already held by 79, therefore it will occupy 5.
As a result, option (C) is the right answer.
4. A random shuffled deck's top and bottom cards have a ______ chance of being both aces.
 4/52 x 4/52
 4/52 x 3/52
 4/52 x 3/51

4/52 x 4/51
Ans. c) 4/52 x 3/51
Explanation: A pack has 52 cards in total.
A pack of four aces has a total of four aces.
The chance of drawing an ace is 4/52.
The chance that both cards are aces is (4/52). (3/51)
5. A sixsided fair die with four green and two red sides is rolled seven times. Which of the following combinations is most likely to be the experiment's outcome?
 Three green faces and four red faces.
 Four green faces and three red faces.
 Five green faces and two red faces.

Six green faces and one red face.
Ans. c) Five green faces and two red faces.
Explanation:
Probability of getting green faces = 4/6 = 2/3
Probability of getting red faces = 2/6 = 1/3
Probability of getting 4 red faces and 3 green faces = P(R_{4}, G_{3})
P(R_{4}, G_{3}) = ^{7}C_{4} (â…“)^{4} * (â…”)^{3 }
Probability of getting 3 red faces and 4 green faces = P(R_{3}, G_{4})
P(R_{3}, G_{4}) = ^{7}C_{3} (â…“)^{3} * (â…”)^{4}
Probability of getting 2 red faces and 5 green faces = P(R_{2}, G_{5})
P(R_{2}, G_{5}) = ^{7}C_{2} (â…“)^{2} * (â…”)^{5}
Probability of getting 1 red face and 6 green faces = P(R_{1}, G_{6})
P(R_{1}, G_{6}) = ^{7}C_{1} (â…“)^{1} * (â…”)^{6}
Since the probability of getting 2 red faces and 5 green faces is the highest, Option (c) is correct.
6. P and Q decide to roll two identical dice, each separately with six sides numbered 1 to 6. If there is a tie, the person with the lowest number wins. If there is still a tie, they roll the dice again until there is no tie. A trial is defined as a dice toss between P and Q. Assume that all six numbers on each dice are equally likely and that all trials are separate. The chances of one of them winning on the third trial are ______ (rounded to three decimal places).
 0.6944
 0.1157
 0.023

0.463
Ans. c) 0.023
Explanation: There are two identical dice with six faces apiece. (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) are favourable circumstances for a tie.
Because there are two dice thrown, the total sample space is 6 Ă— 6 = 36.
As a result, P(tie) = 6/36 = 1/6 and P(not tie) = (1  1/6)
The previous two trials should be a draw if one of them wins the third trial = 1/6 * 1/6 * (1  1/6) = 1/36 * 5/6 = 5/216 = 0.023
As a result, option (C) is the right answer.
7. There are 19 red balls and 19 black balls in a bag. Two balls are frequently taken and discarded if they are the same color; if they are different colors, the black ball is discarded, and the red ball is returned to the bag. The chances of this process ending with one red ball are
 1/21
 0
 0.5

1
Ans. d) 1
Explanation: RR, BB, RB, and BR are possible outcomes. If two balls of the same color appear, both are discarded, but if two balls of different colors appear, just the black ball is discarded. Regardless of the experiment's outcome, black balls will always be used up, and just one red ball will remain in the end. There is always a onetoone chance that this process will end with one red ball.
8. What is the probability of two classmates being born in the same month?
 1/6
 1/12
 1/144

1/24
Ans. b) 1/12
Explanation: A person's chance of being born in a certain month is 1/12. Because a year has 12 months, the probability of both friends being born in January is 1/12 * 1/12. Probability of both friends being born in February = 1/12 * 1/12, etc.
Probability of both friends being born in the same month
= 12 * (1/12 * 1/12)
= 1/12
9. A box contains three cards. One card has two black sides, one card has two red sides, and the third card has one black side and one red side. We choose a card at random and look at one side. What is the likelihood that the opposing side will be the same color as the one we saw?
 3/4
 2/3
 1/2

1/3
Ans. 2/3
Explanation: There are three possible outcomes: BB, RR, and BR, for a total of three possible outcomes: the likelihood that the other side is the same color as the one we observed: Because BB and RR will have the same color on the opposite side, the best outcome will be 2. The probability will be 2 / 3 (favorable/total outcome).
10. A class of 30 students sits in a classroom with five rows of chairs, each with eight seats. The likelihood that the sixth seat in the fifth row will be unoccupied if the students seat themselves at random is
 1/5
 1/3
 1/4

2/5
Ans. c) 1/4
Explanation: There are five rows of chairs, each with eight seats. As a result, there are a total of 40 seats available. If the sixth seat in the fifth row is vacant, 30 students will have 39 options.
As a result, the number of ways to pick from a set of options = ^{39}C_{30}.
However, the total number of options is ^{40}C_{30}. Probability = ^{39}C_{30} / ^{40}C_{30}
Probability = 1/4
For more GATE questions on probability, refer to our Gate Questions on probability: Part 1 and Gate Questions on probability: Part 2.