Gate Questions on Probability: Part 2

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 coin-toss coming up heads is 1/2.

We have curated a course just for you to learn and hone your aptitude skills. Learn Probability from Coding Ninjas to take your aptitude game to the next level.

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. To make a random bit string of length n, a fair coin is thrown n times, and the bit is set to 0 or 1 depending on whether the outcome is head or tail. The chances of two such randomly generated strings not being identical is

1. 1/2ⁿ
2. 1 - (1/n)
3. (1/n!)
4. 1 - (1/2ⁿ)
    

Ans. d) 1 - (1/2ⁿ)

Explanation: 

Assume the outcome is head => 0 and tail => 1. 

P(H) = P(T) = 1/2 (since the coin is fair) 

The string's length is n. 

Both strings should not be identical (P(X)). P(-X)

= they aren't the same

= 1 – P (X).

If both the strings are equal, every character should be same w.r.t its positions i.e P(X) = 1/2*1/2*.......(n times) = (1/2)ⁿ 

P(-X) = 1 – (1/2)ⁿ
 

2. If you throw a fair coin four times. What is the likelihood of getting two heads and two tails?

1. 1/2
2. 5/8
3. 2/4
4. 3/8
    

Ans. d) 3/8

Explanation: There are 16 choices, with two heads and two tails for each: HHTT, HTHT, TTHH, THTH, HTTH, THHT. As a result, the likelihood of two heads is 6/16, or 3/8.

 

3. There are 150 one-mark multiple-choice questions on each test paper, each with four options. Each wrong answer is worth -0.25 points. Assume that 1000 students select all of their responses at random with a consistent probability. The total number of predicted marks for all of these pupils is:

1. 0
2. 2550
3. 7525
4. 9375
    

Ans. d) 9375

Explanation: 

Expected marks per question  = -0.25 * 3/4 + 1 * 1/4 = 1/16

Because the choices are evenly distributed, the predicted marks are 150*1000/16 = 9375.

 

4. S1 and S2, two n-bit binary strings, are randomly picked with uniform probability. The probability of Hamming distance between these strings (the number of bit places where the two strings vary) is likely to be equal to d is 

1. ⁿC_d /2ⁿ
2. ⁿC_d / d
3. d/2ⁿ
4. 1/2^d
    

Ans. a) ⁿC_d /2ⁿ

Explanation: There are n binary bits in which only d might differ.

Ways to choose these d bits =ⁿC_d 

the probability of d bits differs, but n-d bits do not =(1/2)^d * (1/2)^(n-d) the number of ways to choose these d bits

No. of way = ⁿC_d /2ⁿ

 

5. Four fair coins are tossed at the same time. The likelihood of at least one head and one tail appearing is:

1. 1/16
2. 1/8
3. 7/8
4. 15/16
    

Ans. c) 7/8

Explanation: There are just two exceptions to the given output (when all heads or all tails). So (16-2)/16 = 7/8 is the necessary probability.
 

6. In a week, there were seven (different) vehicle accidents. What are the chances that they all happened on the same day?

1. 1/7⁷
2. 1/7⁶
3. 1/2⁷
4. 7/2⁷
    

Ans. b) 1/7⁶

Explanation: Prob(all Monday accidents) = 1/7⁷. Similarly, for the remaining six days. As a result, total probability equals 7 * 1/7⁷ = 1/7⁶. As a result, option (B) is accurate.

 

7. Consider a two-team tournament with three games. Assume that each match will end in a victory or a defeat. The series is won by the side that wins two or more games. Both teams have a 1/2 chance of winning the opening match. A team's chances of winning a game following a win are 2/3. After a loss, the chances of winning a game are 1/3. It's worth noting that just the previous game's impact is considered. What are a team's chances of winning the series after losing the first game?

1. 1/9
2. 1/6
3. 2/9
4. 1/3
    

Ans. c) 2/9

Explanation: A team gets three opportunities to win the event (games). A team must win at least two games to win the tournament. According to the question, the team loses the first chance. Now, to win, the remaining two games must be won. The chances of winning the second game are 1/3. (as the previous game was a loss) 2/3 chance of winning the third game (as the previous game was a win). As a result, the probability of winning the event is 1/3 * 2/3 = 2/9.
 

8. Two numbers, one from each set, are chosen at random from Set A = 2, 3, 4, 5 and Set B = 11, 12, 13, 14, 15. How likely is it that the total of the two integers equals 16?

1. 0.20
2. 0.25
3. 0.30
4. 0.33
    

Ans. a) 0.20

Explanation: There are 20 potential pairs based on the numbers 2, 3, 4, 5 and 11, 12, 13, 14, 15 (i.e. 5*4=20). 

The summation of only the following pairs (4) is 16.

(2, 14)

(3, 13)

(4, 12)

(5, 11)

Therefore, Probability = 4/20 = 0.20

 

9. How many different ways can we place five different balls, B1, B2,..., B5, in 5 different cells, C1, C2,..., C5, so that Ball B is not in cell C, Vi=1,2,...,5, and each cell has exactly one ball?

1. 44
2. 96
3. 120
4. 312
    

Ans. a) 44

Explanation: Number of ways balls are in different cells = Total outcomes - Number of ways for at least one ball to be in a cell.

The total number of permutations conceivable is 5!=120.

We now calculate the number of non-derangements. For ball 1 at position one, we have the rest of the combinations as 4!. For ball 2 at cell 2, we have the rest of the combinations as 4! And so on.

Therefore the number of ways for the ith ball to be in their respective ith cell = 5C1*4!

Similarly, the number of ways for ith and i+1th ball to be in their respective cells = 5C2*3!

Therefore, 

Number of possible ways for at least one ball to be in a cell = 5C1*4! - 5C2*3! + 5C3*2! + 5C4*1! + 1 = 76 

Hence, the number of ways balls are in different cells = 120-76=44

 

10. There are ten blue marbles, 20 green marbles, and 30 red marbles in each bag. A marble is withdrawn from the bag, its color is noted, and returned to the bag. This procedure is performed three times more. What's the chance that no two of the marbles selected will be the same color?

1. 1/36
2. 1/6
3. 1/4
4. 1/3
    

Ans. b) 1/6 

Explanation: Because there are three colors, the number of possible combinations is 3! = 6. Blue marble's chances are 10/60. Green and red marble have a 20/60 and 30/60 chance of appearing, respectively. The probability that no two marbles are picked of the same color equals 6 * (10/60 * 20/60 * 30/60) = 1/6.

 

For more GATE questions on probability, refer to our Gate Questions on probability: Part 1 and Gate Questions on probability: Part 3.