Discrete Mathematics Probability GATE Previous Year Solved Questions

Ques 1: Consider Guwahati, (G) and Delhi (D)whose temperatures can be classified as high (H), medium (M) and low (L). Let P(H_G) denote the probability that Guwahati has high temperature. Similarly, P(M_G) and P(L_G)denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(H_D),P(M_D) and P(L_D) for Delhi.

GATE 2018

The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.

Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (H_G) then the probability of Delhi also having a high temperature (H_D) is 0.40; i.e., P(H_D∣H_G)=0.40 Similarly, the next two entries are P(M_D∣H_G)=0.48 and P(L_D∣H_G)=0.12. Similarly for the other rows.

If it is known that P(H_G)=0.2,P(M_G)=0.5, and P(L_G)=0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is ________.

Ans: 0.60

Solution: P(H_G/H_D)=P(H_G H_D)/P(H_D)=0.2*0.4/(0.2*0.4+)

Ques 2: Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is ____

GATE 2018

Ans: 0.023

Solution: Here, NOT SUCCESS, p=6/36=1/6

SUCCESS, q=30/36=5/6

So the required case is ppq=p²q

=(1/6)x(1/6) x(5/6)

=5/216

=0.023

Ques 3: If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X+2)²] equals ___.

GATE 2017 Set 2

Ans: 54

Solution: For poisson distribution, mean and variance are same.

So, E(X)=V(X)=5

Also, V(X)=E(X²)-(E(X))²

5=E(X²)-25

E(X²)=30

Now, E(X+2)²

=E(X²+ 4X +4)

=E(X²) + E(4X) +4

=E(X²) + 4E(X) +4

=30+4(5)+4

=30+20+4

=54

Ques 4: P and Q are considering to apply for a job. The probability that P applies for the job is1/4, the probability that P applies for the job given that Q applies for the job is 1/2, and the probability that Q applies for the job given that P applies for the job is 1/3. Then the probability that P does not apply for the job given that Q does not apply for this job is

GATE 2017 Set 2

(A) 4/5

(B) 5/6

(C) 7/8

(D) 11/12

Ans: (A) 4/5

Solution: Let, P= P applies for the job

Q= Q applies for the job

---

Given, P(P)=1/4  

P(P’)=1-1/4=3/4

Given, P(Q∣P)=1/3

P(P⋂Q)/P(P)=1/3

P(P⋂Q)/(1/4)=1/3

P(P⋂Q)=(1/3)X(1/4)=1/12

Given, P(P∣Q)=1/2

=P(P⋂Q)/P(Q)=1/2

(1/12)/P(Q) =1/2

P(Q)=(1/12)/(1/2)=1/6

P(Q’)=1-1/6=5/6

Now, P(P’⋂Q’)

=P(PUQ)’

=1-P(PUQ)

=1-(P(P) + P(Q)- P(P⋂Q))

1-(1/4 + 1/6 -1/12)

=1-1/3

=2/3

Now, we need to find P(P′∣Q′)

=P(P’⋂Q’)/P(Q’)

=(2/3) /(5/6)

=4/5

Correct Answer: A

Ques 5: Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100100 hours given that it is of Type 11 is 0.70.7, and given that it is of Type 22 is 0.40.4. The probability that an LED bulb chosen uniformly at random lasts more than 100100 hours is _________.

GATE 2016

Ans: 0.55

Solution: Given that the shop has an equal number of LED bulbs of two different types. Therefore,

Probability of Taking Type 1 Bulb =0.5

Probability of Taking Type 2 Bulb =0.5

The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. i.e.,

Prob(100+∣Type1)=0.7

Prob(100+∣Type2)=0.4

Prob(100+)=Prob(100+∣Type1)×Prob(Type1)+Prob(100+∣Type2)×Prob(Type2)

=0.7×0.5+0.4×0.5=0.55

Ques 6: Consider the following experiment.

GATE 2016 Set 1

Step 1. Flip a fair coin twice.

Step 2. If the outcomes are (TAILS, HEADS) then output YY and stop.

Step 3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output NN and stop.

Step 4. If the outcomes are (TAILS, TAILS), then go to Step 1.1.

The probability that the output of the experiment is Y is————– (up to two decimal places)

Ans: 0.33

Solution: 1st time it is (1/4).

When tail comes, entire process gets repeated

So next time probability of Y to happen is (1/4×1/4),

Likewise it goes on as infinite GP

Sum of infinite GP =a/(1−r)

=(1/4)/(1-1/4)

=(1/4)/(3/4)

=1/3

=0.33

Ques 7: A probability density function on the interval [a,1] is given by 1/x² and outside this interval the value of the function is zero. The value of a is _________.

GATE 2016 Set 1

Ans: 0.5

Solution: f(x)=1/x2 a≤x≤1

=0 Otherwise

Now We know that the sum of all the probabilities is 1

Therefore, on integrating 1/x² with limits a to 1, the result should be 1.

Hence,  ∫1/x² dx=1

            [-1/x]¹_a =1

-1 + 1/a =1

1/a=2

a=1/2            

Hence, a = 0.5

Ques 8: Let S be a sample space and two mutually exclusive events A and B be such that A∪B=S. If P(.) denotes the probability of the event, the maximum value of P(A)P(B) is_____.

GATE 2014 Set 3

Ans: 0.25

Solution: Since both are mutually exclusive P(A⋂B)=0

A∪B=S

P(A U B )=P(S)

P(A)+P(B) + P (A⋂B) =1

P(A)+P(B) =1

When sum is a constant, product of two numbers becomes maximum when they are equal.

So,P(A)=P(B)=1/2

P(A) x P(B)=(1/2)x (1/2) =1/4 =0.25

Ques 9: The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______ .

GATE 2014 Set 2

Ans: 0.26

Solution:Number of integers divisible by 2=50

Number of integers divisible by 3=33

Number of integers divisible by 5=20

Number of integers divisible by 2 and 3= 16

Number of integers divisible by 2 and 5=10

Number of integers divisible by 3 and 5=6

Number of integers divisible by 2 and 3 and 5=3

Total numbers divisible by 2 or 3 or 5=50+33+20−16−10−6+3=74

Total number not divisible by 2 or 3 or 5=26

Required Probability=26/100=0.26

Ques 10: The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p=_____________.

GATE 2014 Set 2

Ans:11.90

Solution: Initially P (working computer) =4/10

P (non-working computer) =6/10

P(atleast 3)= P(3) + P(4)

Case 1 : Three computers are functional

There are 4 sub cases WWWN, WWNW, WNWW, NWWW

where W means working, N means non-working,

but P(WWWN)=P(WWNW)=P(WNWW)=P(NWWW)

P(WWWN)=4/10×3/9×2/8×6/7=144/5040

In all other 3sub-cases, we get same numerators and denominators (in different order), so total prob in this case is

=(4×144)/5040=576/5040

Case 2 : all 4 are working

P(WWWW)=4/10×3/9×2/8×1/7=24/5040

So, P(atleast 3 are working) =576/5040 + 24/5040

=600/5040= 60/504

So 100×p=6000/504

=11.90