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 when it comes to 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 2digit number must be chosen randomly from all the 2digit integers between 1 and 100. What is the probability that the chosen integer is not divisible by seven?
 12/90
 78/90
 77/90

13/90
Ans. c) 77/90
Explanation: There are 90 twodigit numbers in total, 13 of which are divisible by 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, and 98. As a result, the probability that the chosen number is not divisible by 7 is 1  13/90 = 77/90. As a result, (C) is the correct answer.
2. Consider a random variable X that has a probability of 0.5 for each of the values +1 and 1. At x = 1 and +1, the values of the cumulative distribution function F(x) are
 0 and 0.5
 0 and 1
 0.5 and 1

0.25 and 0.75
Ans. c) 0.5 and 1
Explanation: The Cumulative Distribution Function (CDF) is a mathematical function that is calculated as G(x) = P(X≤x)
G(1) = P(X≤1) = P(X=1) = 0.5
G(+1) = P(X≤+1) = P(X=1) + (P=+1) = 0.5+0.5 = 1
3. When two fair coins are flipped, and at least one of the outcomes is known to be a head, what are the chances of receiving both heads?
 1/3
 1/4
 1/2

2/3
Ans. a) 1/3
Explanation: We only have three choices because we know one of the outcomes is head, {h, t}, {h, h}, {t, h}. Both heads have a 1/3 chance of appearing.
4. What is the difference between the square of the expectation of a random variable (E[X2]) and the square of the expectation of the random variable (E[X])2?
 R = 0
 R < 0
 R >= 0

R > 0
Ans. c) R >= 0
Explanation: Variance of a random variable refers to the difference between (E[X2]) and (E[X])2. The variance of a group of integers indicates how widely they are dispersed. (A variance of 0 means that all the values are the same.) Variance is always positive when it is not zero
5. A deck of five cards (each bearing a different number from 1 to 5) is thoroughly shuffled. The deck is then emptied of two cards at a time. What is the chance that the two cards are chosen with the first card's number being one higher than the second card's number?
 1/5
 4/25
 1/4

2/5
Ans. a) 1/5
Explanation: We must choose two cards from a deck of five. Because the sequence in which they are drawn is important, there are 5P2 = 5!/3! = 20 primary occurrences from which there are four favorable cases: 5 comes before 4, 4 comes before 3, 3 comes before 2, and 2 comes before 1. As a result, probability equals 4/20 = 1/5.
6. Consider a computerassembly business. The likelihood of a malfunctioning computer assembly is p. As a result, each computer is placed through a rigorous testing procedure. For any computer with a probability of q, this testing technique yields the correct result. What are the chances of a computer being labeled defective?
 (1  q) p
 (1  p) q
 pq

pq + (1  p)(1  q)
Ans. d) pq + (1  p)(1  q)
Explanation:
In two situations, a computer can be labeled faulty.
Source: https://testbook.com/questionanswer/consideracompanythatassemblescomputersthep60b62f9ed929e08081c85c84
1) It is, in fact, defective and has been accurately declared as such (p*q).
2) Not defective and declared wrongly (1p)* (1q).
The chances of a computer being labeled defective = pq + (1  p)(1  q)
7. Every day, Aishwarya learns either computer science or mathematics. If she studies computer science one day, she has a 0.6 probability of studying mathematics the following day. If she studies mathematics one day, she has a 0.4 chance of studying computer science the next. What is the probability that Aishwarya will study computer science on Wednesday, given that she studies computer science on Monday?
 0.24
 0.36
 0.4

0.6
Ans. c) 0.4
Explanation: Aishwarya studies computer science on Monday >. Her chances of studying mathematics on Tuesday are 0.6, while her chances of studying computer science on Tuesday are 0.4 —> On Tuesday, she studies mathematics. On Wednesdays, she studies computer science = 0.6 x 0.4 = 0.24. On Tuesday, she studies computer science, and on Wednesday, she studies computer science = 0.4x0.4 = 0.16. By adding 1 and 2, the probability that she will study computer science on Wednesday is 0.24 + 0.16 = 0.40.
Source: https://gateoverflow.in/425/Gatecse2008question27
8. Let's say we pick a uniform and random permutation from the 20! What's the chance that two comes up first in the chosen permutation before any other even number?
 1/2
 1/10
 9!/20!

None of the above
Ans. b) 1/10
Explanation: The probability of being first is the same for all even values (The odd numbers do not matter here). There are a total of ten of them. The chances of 2 coming before the other nine evens are 1/10. Therefore, option (B) is the correct answer.
9. Assume you break a unitlength stick at an evenly chosen random spot. The shorter stick's anticipated length is
 0.24 to 0.27
 0.15 to 0.30
 0.20 to 0.30

0.10 to 0.15
Ans. a) 0.24 to 0.27
Explanation: The smaller sticks will be available in lengths ranging from nearly 0 units to 0.5 units, with any length equally conceivable. As a result, the average length is approximately (0 + 0.5)/2 = 0.25 unit, which is the correct answer.
10. An unbiased coin is tossed for each element in a set of size 2n. The tosses of the 2n coins are random. An element is picked if the appropriate coin flip results in a head. The chance of selecting exactly n components is:
 (^{2n}C_{n}) / (4^{n})
 (^{2n}C_{n}) / (2^{n})
 1 / (^{2n}C_{n})

1/2
Ans. a) (^{2n}C_{n}) / (4^{n})
Explanation: The major focus of the question is on the probability of n heads out of 2n coin tosses.
P = ^{2n}C_{n}∗((1/2)^{n})∗((1/2)^{n})
= (^{2n}C_{n}) / (4^{n})
For more GATE questions on probability, refer to our Gate Questions on probability: Part 2 and Gate Questions on probability: Part 3