Introduction
Probability, in general terms, means the possibility of an event occurring. Probability is a branch of mathematics that tells us the chances that a given event will occur. The probability for an impossible event is 0 and for a sure event is 1.
Problems
Now, we will see the Probability Previous Year's Questions asked in the GATE exam.

A 2digit number is selected randomly out of all 2digit integers between 1 and 100. What will be the probability that the number selected is not divisible by 7?
A) 13/90
B) 77/90
C) 56/90
D) 79/90
ANSWER: B
Explanation: The total number of twodigit numbers between 1 and 100 is 90. Out of the 13 are found to be divisible by 7, are 14,21,28,35,42,49,56,63,70,77,84,91 and 98. .So, the probability that the selected number is divisible by 7 is 13/90. This means the probability that the selected number is not divisible by 7 is 113/90, i.e., 77/90.Option(B) is correct

A fair sixsided die is rolled once. If the value on the die is found to be 1,2,3, we will roll the die for a second time. What would be the probability that the sum of values turns up at least 6?
A) 5/12
B)10/21
C)3/7
D)2/3
ANSWER: A
Explanation: For the following, even sum would be greater than or equal to 6.6 appeared on the first throw (â…™)+1 appeared on the first throw, 5 appeared on the second throw (â…™)*(â…™) + 1 appeared on the first throw, and 6 on the second (â…™)*(â…™). Similarly, if 2 appeared on the first throw, and four appeared on the second throw(â…™)*(â…™), + 2 appeared on the first throw, and 5 appeared on the second throw(â…™)*(â…™)+ 2 appeared on the first throw, and 6 appeared on the second throw(â…™)*(â…™) + and similarly, if 3 appeared on the first throw, and three appeared on the second throw(â…™)*(â…™), + 3 appeared on the first throw, and 4 appeared on the second throw(â…™)*(â…™)+ 3 appeared on the first throw, and 5 appeared on the second throw(â…™)*(â…™) + 3 appeared on the first throw, and 6 appeared on the second throw(â…™)*(â…™) So P(x) = â…™+ (â…™)(â…™)+(â…™)(â…™)+(â…™)(â…™)+(â…™)(â…™)+(â…™)(â…™)+(â…™)(â…™)+(â…™)(â…™)+(â…™)(â…™)+(â…™)(â…™)= 1/6+9/36= 5/12
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Consider a random variable X that takes value +1 and 1 with fair probability of 0.5 each.What would be the cumulative distribution function F(x) at x = 1 and +1 are
A) 0 and 0.5
B) 0 and 1
C) 0.5 and 1
D)0.5 and 0.75
ANSWER: C
Explanation :The Cumulative distribution function F(x) = P(Xâ‰¤x) F(1) = P(Xâ‰¤1) = P(X=1) = 0.5 F(+1) = P(Xâ‰¤+1) = P(X=1) + (P=+1) = 0.5+0.5 = 1

Two fair coins are flipped, and it's known that at least one of the outcomes is a head. What would be the probability that both outcomes are head?
A)1/2
B)1/4
C)1/3
D)2/3
ANSWER: C
Explanation: Since it's known that at least one of the outcomes is a Head, there remain only three possibilities that are (H, H), (T, H), (H, T). So The Probability of both heads = 1/3.

What is the difference between the expectation of the square or a random variable E[XÂ˛] and the square of the expectation of a random variable (E[X])Â˛.
A)The difference is equal to 0
B)The difference is greater than 0
C)The difference is lesser than 0
D)The difference is greater than equal to 0
ANSWER: D
Explanation: The difference between the (E[XÂ˛]) and (E[X])Â˛ is called the variance of a random variable. (If the variance is zero, all the values are identical.) A nonzero variance is always positive.

In a deck of 5 cards, each card carrying a specific number from 1 to 5 is shuffled thoroughly. Two cards are removed from the deck one at a time. What would be the probability that two cards are selected such that the number on the first card is found to be one higher than the number on the second card?
A) 4/25
B) 1/5
C) 1/4
D) 2/5
ANSWER: B
Explanation: We have to select two cards from the deck of 5 Cards. Since the order in which cards are drawn matters here, there are 5P2 = 5!/3! i.e. 20 elementary events out of which only 4 are favourable cases: 5 comes before 4, 4 comes before 3, 3 comes before 2 and 2 comes before 1. Hence, probability = 4/20 = 1/5

What will be the probability that a divisor of 1099 is found to be a multiple of 1096?
A)1/625
B)4/625
C)8/625
D)12/625
ANSWER: A
Explanation :Multiples of 1096 that are also the divisor of 1099 are 1096, 2x1096, 4x1096, 5x1096, 8x1096, 10x1096, 20x1096, 25x1096, 40x1096, 50x1096, 100x1096, 125x1096, 200x1096, 250x1096, 500x1096, 1000x1096 The total number of divisors of 1099 = 10000 So the probability = 16/10000 = 1/625

Mathur studies either physics or chemistry every day. If she studies physics on one day, then the probability that she studies chemistry the next day is 0.6. If she studies chemistry on one day, then the probability that she studies physics the next day is 0.4. Given that Mathur studies physics on Monday, what is the probability that he studies physics on Wednesday?
A)0.24
B)0.32
C)0.36
D)0.40
ANSWER: D
Explanation: Mathur studies physics on Monday. Then the probability that he studies chemistry on Tuesday is 0.6, and the probability that he studies physics on Tuesday is 0.4. He studies chemistry on Tuesday and physics on Wednesday = 0.6 x 0.4 = 0.24 or He studies physics on Tuesday and physics on Wednesday = 0.4x0.4 = 0.16 Adding 1 and 2, the required probability that he studies physics on Wednesday is 0.24 + 0.16 = 0.40

If you broke a stick of unit length at any point chosen uniformly at random. What would b the expected length of the shorter part of the stick
A)0.24 to 0.27
B)0.15 to 0.30
C)0.20 to 0.30
D)0.10 to 0.15
ANSWER: A
Explanation: If we broke a stick of unit length at any point chosen uniformly at random. The smaller part of the stick would range in length from almost 0 units up to a maximum of 0.5 units, where each length is equally possible. Hence, the average length will be about (0 + 0.5)/2 = 0.25 unit.

Suppose four fair sixsided dice are rolled. What would be the probability that the sum of the number in dice being 22
A) 7/1296
B) 8/1296
C) 9/1296
D) 10/1296
ANSWER: D
Explanation: Probability (of an event ) is defined as No of favourable outcomes to the event / Total Number of possible outcomes in the random experiment. Here, four sixfaced dice are rolled. There could be six equally likely and mutually exclusive outcomes for one dice. Taking four together, there can be a total number of 6*6*6*6 = 1296 possible outcomes. There are no favourable cases to the event: here, the event is getting a sum of 22. So, there can be only 2 possible casesâ€”case 1: Three 6's and one 4, for example, 6,6,6,4 ( sum is 22). Hence, the number of ways we could obtain this is 4!/3! i.e., four ways ( 3! is for removing those cases where all three six are swapping among themselves) Second Case has two 6's and two 5's, for example, 6,6,5,5 that also sum to 22. Therefore the number of ways we can obtain this = 4! /( 2! * 2!) = six ways ( 2! is for removing those cases where both six are swapping between themselves, similarly for both five also). Hence total no of favourable cases = 4 + 6 = 10. Hence probability = 10/1296. Therefore option D.

What will be the probability of having two heads and two tails if a fair coin is tossed four times
A) 3/8
B) 5/8
C) 1/2
D) 3/4
ANSWER: A
Explanation: If a coin is tossed four times, there are a total of 16 possibilities, out of which only the following mentioned have occurrence two heads and two tails, i.e., HHTT, HTHT, HTTH, TTHH, THTH, THHT So the Probability of having two heads is 6/16 which is â…ś

Suppose four fair coins are tossed simultaneously. What will be the probability that at least one head and one tail will turn up is:
A)1/16
B)1/8
C)7/8
D)15/16
ANSWER: C
Explanation: The probability that at least one head or at least one tail would appear will be 1 the probability that no head or no tail appears. Since there are only two cases for which only head r only tail appears are TTTT, HHHH so the required probability is 11/8 = â…ž

If seven distinct car accidents occur in a week. What would be the probability that all distinct accident happens on the same day of the week
A) 1/(7^7)
B) 1/(7^6)
C) 1/(2^7)
D) 7/(2^7)
ANSWER: B
Explanation: The probability of all accidents happening on Monday is 1/(7^7). Similarly, for the rest of the six days. So the total probability that all seven accident happens on the same day is 7*(1/7^7) = 1/(7^6)

Given two Set P = {2, 3, 4, 5} and Set Q = {11, 12, 13, 14, 15}, two numbers are selected at random, one from each set. What will be the probability that the sum of the two numbers equals 16?
A)0.20
B)0.25
C)0.30
D)0.35
ANSWER: A
Explanation :There are 20 possible pairs we could draw from Set P {2, 3, 4, 5} and Set Q {11, 12, 13, 14, 15} i.e 5*4=20 Out of which following pairs have sum equals to 16 (2, 14),(3, 13),(4, 12),(5, 11) Probability is equal to No. Favourable Outcomes/Total No. Outcomes Probability = 4/20 = 0.20

A bag contains ten blue balls, 20 green balls and 30 red balls. A ball is drawn from the bag, and its colour is recorded and put back in the bag again. This process is repeated three times. What will be the probability that no two of the balls drawn have the same colour is
A)1/6
B)1/4
C)1/36
D1/3
ANSWER: A
Explanation: As the number of the ball's colour is 3, Possible combinations would be 3! = 6. Probability of Blue ball: 10/60 Probability of Green ball: 20/60 Probability of Red ball: 30/60.The probability that no two of the balls drawn have the same colour is = 6 * (10/60 * 20/60 * 30/60) = 1/6