Question 1
Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
A
8 / (2e3)
B
9 / (2e3)
C
17 / (2e3)
D
26 / (2e3)
GATE CS 2013    Probability    
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Question 1 Explanation: 
See http://en.wikipedia.org/wiki/Poisson_distribution#Definition
PR(X < 3) = Pr(x = 0) + Pr(x = 1) + Pr(x = 2) = f(0, 3) + f(1, 3) + f(2, 3) Put [Tex]\lambda[/Tex] = 3 and k = 0, 1, 2 in the formula given at http://en.wikipedia.org/wiki/Poisson_distribution#Definition = 17 / (2e3)
Question 2
Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7?
A
13/90
B
12/90
C
78/90
D
77/90
GATE CS 2013    Probability    
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Question 2 Explanation: 
There are total 90 two digit numbers, out of them 13 are divisible by 7
Question 3
Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
A
10/21
B
5/12
C
2/3
D
1/6
GATE CS 2012    Probability    
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Question 3 Explanation: 
The following are different possibilities (1,5) (1,6) (2,4) (2,5) (2,6) (3,3) (3,4) (3,5) (3,6) Plus 1/6 probability that first time 6 is rolled So total probability is 9/36 + 1/6 = 15/36 = 5/12.
Question 4
Consider a random variable X that takes values +1 and −1 with probability 0.5 each. The values of 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
0.25 and 0.75
GATE CS 2012    Probability    
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Question 4 Explanation: 
The Cumulative Distribution Function F(x) = P(X≤x) F(-1) = P(X≤-1) = P(X=-1) = 0.5 F(+1) = P(X≤+1) = P(X=-1) + (P=+1) = 0.5+0.5 = 1
Question 5
If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads?
A
1/3
B
1/4
C
1/2
D
2/3
GATE CS 2011    Probability    
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Question 5 Explanation: 
Since we know one outcome is head, there are only three possibilities {h, t}, {h, h}, {t, h} The probability of both heads = 1/3
Question 6
If the difference between expectation of the square of a random variable (E[X²]) and the square of the expectation of the random variable (E[X])² is denoted by R, then?
A
R = 0
B
R < 0
C
R >= 0
D
R > 0
GATE CS 2011    Probability    
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Question 6 Explanation: 
The difference between  (E[X²]) and (E[X])² is called variance of a random variable.  Variance measures how far a set of numbers is spread out. (A variance of zero indicates that all the values are identical.) A non-zero variance is always positive:
Question 7
Consider a finite sequence of random values X = { x1, x2,..., xn}. Let μx be the mean and σx be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as yi = a*xi + b, where a and b are positive constants. Let μy be the mean and σy be the standard deviation of this sequence. Which one of the following statements is INCORRECT?
A
Index position of mode of X in X is the same as the index position of mode of Y in Y.
B
Index position of median of X in X is the same as the index position of median of Y in Y.
C
μy = aμx+b
D
σy = aσx+b
GATE CS 2011    Probability    
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Question 7 Explanation: 
Adding a constant like b shift the distribution while multiplying to a constant like a stretch the distribution along median gate2011A29 Mode is the most frequent data of the distribution, so the index position of the mode will not change. From the above graph it is clear that index position of the median will also not change. Now for the mean gate2011A29b And for the standard deviation gate2011A29c     Source: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html
Question 8
A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?
A
1/5
B
4/25
C
1/4
D
2/5
GATE CS 2011    Probability    
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Question 8 Explanation: 
You have to select 2 cards from 5. Since the order in which they are drawn matters, there are 5P2 = 5!/3! = 20 elementary events from which there are 4 favorable number of cases: 5 before 4, 4 before 3, 3 before 2 and 2 before 1. Hence, probability = 4/20 = 1/5   Source: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html
Question 9
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process.This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty?
A
pq + (1 - p)(1 - q)
B
(1 - q) p
C
(1 - p) q
D
pq
GATE CS 2010    Probability    
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Question 9 Explanation: 
A computer can be declared faulty in two cases 
1) It is actually faulty and correctly declared so (p*q)
2) Not faulty and incorrectly declared (1-p)*(1-q). 
Question 10
What is the probability that divisor of 1099 is a multiple of 1096?
A
1/625
B
4/625
C
12/625
D
16/625
GATE CS 2010    Probability    
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Question 10 Explanation: 
Following are multiples of 1096 which are also divisor of 1099 1096, 2x1096, 4x1096, 5x1096, 8x1096, 10x1096, 20x1096, 25x1096, 40x1096, 50x1096, 10x1096, 125x1096, 200x1096, 250x1096, 500x1096, 1000x1096 The total number of divisors of 1099 = 10000  (See http://www.math.cmu.edu/~mlavrov/arml/13-14/number-theory-09-29-13.pdf) So the probability = 16/10000=1/625
Question 11
An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3?
A
0.453
B
0.468
C
0.485
D
0.492
GATE-CS-2009    Probability    
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Question 11 Explanation: 
Let Xi be the probability that the face value is i.

We can say that X1 + X2 + X3 + X4 + X5 + X6 = 1

It is given that  (X1 + X3 + X5) = 0.9*(X2 + X4 + X6)

It is also given that X2 = X4 = X6

Also given that (X4 + X6)/(X4 + X5 + X6) = 0.75

Solving the above equations, we can get X3 as 0.468 
Question 12
Aishwarya studies either computer science or mathematics everyday. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
A
0.24
B
0.36
C
0.4
D
0.6
Probability    GATE CS 2008    
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Question 12 Explanation: 
113
Question 13
Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean -1 and variance unknown If P(X <=-1) = P(Y >-2). the standard deviation of Y is
A
3
B
2
C
sqrt(2)
D
1
Probability    GATE CS 2008    
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Question 13 Explanation: 
115
Question 14
Suppose we uniformly and randomly select a permutation from the 20! Permutations of 1, 2, 3 ,…..,20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?
A
1/2
B
1/10
C
9!/10!
D
Node of the above
Probability    GATE-CS-2007    
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Question 15
We are given a set X = {x1, .... xn} where xi = 2i. A sample S ⊆ X is drawn by selecting each xi independently with probability pi = 1/2. The expected value of the smallest number in sample S is:
A
1/n
B
2
C
sqrt(n)
D
n
Probability    GATE-CS-2006    
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Question 15 Explanation: 
E = (1/(2^1))*(2^1) + (1/(2^2))*(2^2) + … (1/(2^n))*(2^n) = 1+1+…1 (n times addition of 1) = n
Question 16
Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________
A
0.24 to 0.27
B
0.15 to 0.30
C
0.20 to 0.30
D
0.10 to 0.15
Probability    GATE-CS-2014-(Set-1)    
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Question 17
Four fair six-sided dice are rolled. The probability that the sum being 22 is X/1296. The value of X is ________
A
7
B
8
C
9
D
10
Probability    GATE-CS-2014-(Set-1)    
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Question 17 Explanation: 
In general, Probability (of an event ) = No of favorable outcomes to the event / Total number of possible outcomes in the random experiment. Here, 4 six-faces dices are tossed, for one dice there can be 6 equally likely and mutually exclusive outcomes. Taking 4 together, there can be total number of 6*6*6*6 = 1296 possible outcomes. Now, No of favorable cases to the event : here event is getting sum as 22. So, there can be only 2 cases possible. Case 1: Three 6's and one 4, for example: 6,6,6,4 ( sum is 22) Hence, No of ways we can obtain this = 4!/3! = 4 ways ( 3! is for removing those cases where all three 6 are swapping among themselves) Case 2: Two 6's and two 5's,for example: 6,6,5,5 ( sum is 22) Hence, No of ways we can obtain this = 4! /( 2! * 2!) = 6 ways ( 2! is for removing those cases where both 6 are swapping between themselves, similarly for both 5 also) Hence total no of favorable cases = 4 + 6 = 10. Hence probability = 10/1296. Therefore option D.
Question 18
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 = _____________.
A
11.85
Probability    GATE-CS-2014-(Set-2)    
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Question 18 Explanation: 
Total ways to pick 4 computers = 10*9*8*7

Total ways that at least three computers are fine = 
        Total ways that all 4 are fine + Total ways any 3 are fine

Total ways that all 4 are fine = 4*3*2*1

Total ways three are fine = 1st is Not working and other 3 working + 
                            2nd is Not working and other 3 working + 
                            3rd is Not working and other 3 working + 
                            4th is Not working and other 3 working + 
                         = 6*4*3*2 + 4*6*3*2 + 4*3*6*2 + 4*3*2*6
                         = 6*4*3*2*4


The probability = Total ways that at least three computers are fine /  
                  Total ways to pick 4 computers 
                =  (4*3*2*1 + 6*4*3*2*4) / (10*9*8*7)
                = (4*3*2*25) / (10*9*8*7)
                = 11.9% 
Question 19
Each of the nine words in the sentence ”The quick brown fox jumps over the lazy dog” is written on a separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place.)
A
3.8 to 3.9
Probability    GATE-CS-2014-(Set-2)    
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Question 19 Explanation: 
There are total 9 words in the sentence. To find expected length first we count total length of the sentence i.e 35. Now to find expected length divide total length by total words i.e 35/9 so answer is 3.8 or 3.9 Another Explanation: Expected value = ∑( x * P(x) ) = 3*4/9 + 4*2/9 + 5*3/9 = 35/9 = 3.9
Question 20
The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______ .
A
0.259
B
0.459
C
0.325
D
0.225
Probability    GATE-CS-2014-(Set-2)    
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Question 20 Explanation: 
There are total 100 numbers, out of which 

50 numbers are divisible by 2, 
33 numbers are divisible by 3,
20 numbers are divisible by 5

Following are counted twice above
16 numbers are divisible by both 2 and 3
10 numbers are divisible by both 2 and 5
6 numbers are divisible by both 3 and 5

Following is counted thrice above
3 numbers are divisible by all 2, 3 and 5
So total numbers divisible by 2, 3 and 5 are = = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 103 - 29 = 74 So probability that a number is number is not divisible by 2, 3 and 5 = (100 - 74)/100 = 0.26
Question 21
Let S be the sample space and two mutually exclusive events A and B be such that A U B = S. If P(.) denotes the probability of the event. The maximum value of P(A)P(B) is ______
A
0.5
B
0.25
C
0.225
D
0.125
Probability    GATE-CS-2014-(Set-3)    
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Question 21 Explanation: 
Sample Space(S) - A set of all possible outcomes/events of a random experiment. Mutually Exclusive Events - Those events which can't occur simultaneously.   P(A)+P(B)+P(A∩B)=1   Since the events are mutually exclusive, P(A∩B)=0. Therefore, P(A)+P(B)=1   Now, we now that AM >= GM So, (P(A)+P(B))/2 >= sqrt(P(A)*P(B))   P(A)*P(B) <= 1/4
  Hence max(P(A)*P(B)) = 1/4.
  We can think of this problem as flipping a coin, it has two mutually exclusive events ( head and tail , as both can't occur simultaneously). And sample space S = { head, tail }   Now, lets say event A and B are getting a "head" and "tail" respectively. Hence, S = A U B.   Therefore, P(A) = 1/2 and P(B) = 1/2.   And, P(A).P(B) = 1 /4 = 0.25.   Hence option B is the correct choice.
Question 22
For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is:
A
(2nCn) / (4^n)
B
(2nCn) / (2^n)
C
1 / (2nCn)
D
1/2
Probability    GATE-CS-2006    
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Question 22 Explanation: 
The question is mainly about probability of n heads out of 2n coin tosses. P = 2nCn∗((1/2)^n)∗((1/2)^n) = (2nCn) / (4^n)
Question 23
Let f(x) be the continuous probability density func­tion of a random variable X. The probability that a < X ≤ b, is
		 
A) f(b - a)
B) f(b) - f(a)
C)  GATECS2005Q12A
D) GATECS2005Q12B
A
A
B
B
C
C
D
D
Probability    GATE-CS-2005    
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Question 23 Explanation: 
  anil_1   This solution is contributed by Anil Saikrishna Devarasetty.
Question 24
Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows:
(i)  Select a box
(ii) Choose a ball from the selected box such that each ball in
     the box is equally likely to be chosen. The probabilities of
     selecting boxes P and Q are (1/3) and (2/3), respectively.  
Given that a ball selected in the above process is a red ball, the probability that it came from the box P is
A
4/19
B
5/19
C
2/9
D
19/30
Probability    GATE-CS-2005    
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Question 24 Explanation: 
The probability of selecting a red ball  = 
              (1/3) * (2/5) + (2/3) * (3/4) = 2/15 + 1/2 
              = 19/30  [Let it be P1]

Probability of selecting a red ball from box P =
             (1/3) * (2/5) = 2/15 [Let it be P2]

Given that a ball selected in the above process is 
a red ball, the probability that it came from the 
box P is = P1/P2 = (2/15) / (19/30) = 4/19 
Question 25
A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. The probability that two such randomly generated strings are not identical is
A
1/2n
B
1 - (1/n)
C
(1/n!)
D
1 - (1/2n)
Probability    GATE-CS-2005    
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Question 25 Explanation: 
let us suppose if outcome is head =>0, tail => 1 Since the coin is fare, P(H) = P(T) = 1⁄2 Length of the string is => n P(X) = both the strings should not be identical P(-X) = both are not identical = 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 This solution is contributed by Anil Saikrishna Devarasetty
Question 26
If a fair coin is tossed four times. What is the probability that two heads and two tails will result?
A
3/8
B
1/2
C
5/8
D
2/4
Probability    GATE-CS-2004    
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Question 26 Explanation: 
There are total 16 possibilities, out of which following have 2 heads and 2 tails HHTT, HTHT, TTHH, THTH, HTTH, THHT So the probability of 2 heads is 6/16 which is 3/8
Question 27
An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches -0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probablity. The sum total of the expected marks obtained by all these students is:
A
0
B
2550
C
7525
D
9375
Probability    GATE-CS-2004    
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Question 27 Explanation: 
Expected marks per question is = -0.25 * 3/4 + 1 * 1/4 = 1/16
Since choice is uniformly distributed, expected marks = 150*1000/16 = 9375
Question 28
Two n bit binary strings, S1 and S2, are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is
A
nCd /2n
B
nCd / d
C
d/2n
D
1/2d
Probability    GATE-CS-2004    
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Question 28 Explanation: 
There are two string s1 and s2, so probability of selection of any one is 1/2. Length of string is n bit and hamming distance is d bit so applying binomial distribution- nCd(1/2)^d(1/2)^n - d = nCd(1/2)^d+n-d = nCd /2n
Question 29
A point is randomly selected with uniform probability in the X-Y plane within the rectangle with corners at (0,0), (1,0), (1,2) and (0,2). If p is the length of the position vector of the point, the expected value of p2 is
A
2/3
B
1
C
4/3
D
5/3
Probability    GATE-CS-2004    
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Question 29 Explanation: 
Here minimum value of p can be 0 (if point chosen is (0,0), then length of position vector will be 0), and maximum value can be 5√ when point chosen is (1,2), because that is the farthest point from origin. So p can vary from 0 to 5√. Now we know that [Tex]E(p^2) = \int^\sqrt{5}_0 p^2*P(p)\,dp[/Tex] Since p is a uniform random variable, [Tex]P(p) = \frac{1}{\sqrt{5}-0} = \frac{1}{\sqrt{5}}[/Tex] So [Tex]E(p^2) = \frac{1}{\sqrt{5}}\left[\frac{p^3}{3}\right]^{\sqrt{5}}_0 = \frac{5}{3}[/Tex] So option (D) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2004.html
Question 30
Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A | B) and P(B | A) respectively are
A
1/4, 1/2
B
1/2, 1/14
C
1/2, 1
D
1, 1/2
Probability    GATE-CS-2003    
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Question 30 Explanation: 
Conditional probability measures the probability of an event given that another event has occurred. P(A | B) is Probability of A given B has occurred which is 1. P(B | A) is probability of B given that A has occurred which is 1/2
Question 31
A program consists of two modules executed sequentially. Let f1(t) and f2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by :
		 
A)	f1(t) + f2(t)
B)  GATECS2003Q60A
C) GATECS2003Q60B
D) max {f1(t), f2(t)}
A
A
B
B
C
C
D
D
Probability    GATE-CS-2003    
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Question 31 Explanation: 

We assume the total time to be ‘t’ units and f1 executes for 'x’ units.
Since, f1(t) and f2(t) are executed sequentially. So, f2 is executed for ‘t – x’ units.
We apply convolution on the sum of two independent random variables to get probability density function of the overall time taken to execute the program.
f1(t) * f2(t – x) =   2
 
Thus, option (C) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 32
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is :
A
1/16
B
1/8
C
7/8
D
15/16
Probability    GATE-CS-2002    
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Question 32 Explanation: 
There are only two cases (when all head or all tail) against the given output. So the required probability is (16-2)/16 = 7/8.
Question 33
Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?
A
1/77
B
1/76
C
1/27
D
7/27
Probability    GATE-CS-2001    
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Question 33 Explanation: 
Prob(all accidents on Monday) = 1/77. Similarly for other 6 days. So total probability = 7 * 1/77 = 1/76. So option (B) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2001.html
Question 34
E1 and E2 are events in a probability space satisfying the following constraints:
 
    Pr(E1) = Pr(E2)
    Pr(EI U E2) = 1
    E1 and E2 are independent 
The value of Pr(E1), the probability of the event E1 is
A
0
B
1/4
C
1/2
D
1
Probability    GATE-CS-2000    
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Question 34 Explanation: 
Given Constraints: 1. Pr(E1) = Pr(E2) 2. Pr( E1 U E2) = 1 3. E1 and E2 are independent As we know: Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1 ∩ E2) As E1 and E2 are independent events. (cond.3) So Pr(E1 ∩ E2) = Pr(E1) Pr(E2) Pr(E1) = Pr(E2) (cond.2) let probability of Event E1 = x = prob of E2 So, Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1) Pr(E2) 1 = x + x -x* x (cond. 1) 1=2x-x^2 x^2-2x+1 = 0 (x-1)^2 = 0 x = 1 So, Pr(E1) = Pr(E2) = 1 Thus, option (D) is the answer. Reference : https://people.richland.edu/james/lecture/m170/ch05-rul.html This solution is contributed by Nitika Bansal. Another Solution : E1 and E2 are independent events.
Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1) Pr(E2)
Pr(E1) = Pr(E2) (given)
So,
2 * Pr(E1) – Pr(E1)2 = Pr( E1 U E2)
2 * Pr(E1) – Pr(E1)2 = 1
So, Pr(E1) = Pr(E2) = 1
Thus, option (D) is the answer.
Question 35
An ISP has a link of 100Mbps which is shared by its subscribers. Considering the fact that all of its subscribers are active 50% of the time and the probabilities of being active are independent, the ISP has promised 25 Mbps to its 6 subscribers. What is the probability that any subscriber gets degraded service (less than promised speed).
A
1/32
B
5/16
C
1/2
D
7/64
Probability    GATE-CS-2015 (Mock Test)    
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Question 35 Explanation: 
The probability that more than 4 users are active a time is = (6C5 + 6C6)/26
Question 36
Consider a 3 game tournament between two teams. Assume that every game results in either a win or loss. The team that wins two or more games wins the series. The probability for wining the first game for both teams is 1/2. The probability for a team to win a game after a win is 2/3. The probability of wining a game after a loss is 1/3. Note that the effect of only previous game is considered. What is the probability for a team to win the series after loosing first game.
A
1/9
B
1/6
C
2/9
D
1/3
Probability    GATE-CS-2015 (Mock Test)    
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Question 36 Explanation: 
  A team has 3 chances(games) to win tournament. To win the tournament, a team must win atleast 2 games. Acc to question, Team looses first chances. Now, to win, it is must to win left 2 games. Probability to win the 2nd game : 1/3 (as previous game was a loss) Probability to win the 3rd game : 2/3 (as previous game was a win) So, Probability to win the tournament = 1/3 * 2/3  = 2/9   This solution is contributed by Mohit Gupta.
Question 37
Given Set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is the probability that the sum of the two numbers equals 16?
A
0.20
B
0.25
C
0.30
D
0.33
Probability    GATE-CS-2015 (Set 1)    
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Question 37 Explanation: 
There are 20 possible pairs from {2, 3, 4, 5} and {11, 12, 13, 14, 15} i.e 5*4=20 Out of which following pairs have sum 16.
(2, 14)
(3, 13)
(4, 12)
(5, 11) 

Probability = Favorable Outcomes/Total Outcomes
Probability = 4/20 = 0.20

Therefore option A is correct
Question 38
The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p and c:
1. p + m + c = 27/20 
2. p + m + c = 13/20
3. (p) × (m) × (c) = 1/10 
A
Only relation 1 is true
B
Only relation 2 is true
C
Relations 2 and 3 are true
D
Relations 1 and 3 are true
Probability    GATE-CS-2015 (Set 1)    
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Question 38 Explanation: 
1 - (1 - m) (1 - p) (1 - c) = 0.75               -------(1)
(1 - m)pc +  (1 - p)mc + (1 - c)mp + mpc = 0.5   -------(2)
(1 - m)pc +  (1 - p)mc + (1 - c)mp  = 0.4        -------(3)

From last 2 equations, we can derive mpc = 0.1

After simplifying equation 1, we get.
p + c + m - (mp + mc + pc) + mpc = 0.75        
p + c + m - (mp + mc + pc) = 0.65         -------(4)

After simplifying equation 3, we get
pc + mc + mp - 3mpc = 0.4

Putting value of mpc, we get
pc + mc + mp = 0.7

After putting above value in equation 4, we get
 p + c + m - 0.7 = 0.65
 p + c + m = 1.35 = 27/20 
Question 39
Suppose Xi for i = 1, 2, 3 are independent and identically distributed random variables whose probability mass functions are Pr[Xi = 0] = Pr[Xi = 1] = 1/2 for i = 1, 2, 3. Define another random variable Y = X1 X2 ⊕ X3, where ⊕ denotes XOR. Then Pr[Y = 0 ⎪ X3 = 0] = ____________.
A
0.75
B
0.50
C
0.85
D
0.25
Probability    GATE-CS-2015 (Set 3)    
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Question 39 Explanation: 
 

P (A|B) = P (A∩B) / P (B)

P (Y=0 | X3=0) = P(Y=0 ∩X3=0) / P(X3=0)

P(X3=0) = 1⁄2

Y = X1X2 ⊕ X3

The number of possibilities where Y = 0 can be obtained by constructing a table anil_47

From the above table, P(Y=0 ∩X3=0) = 3/8 And P (X3=0) = 1⁄2

P (Y=0 | X3=0) = P(Y=0 ∩X3=0) / P(X3=0) = (3/8) / (1/2) = 3⁄4 = 0.75

This solution is contributed by Anil Saikrishna Devarasetty .

Another Solution :

It is given X3 = 0. Y can only be 0 when X1 X2 is 0. X1 X2 become 0 for X1 = 1, X2 = 0, X1 = X2 = 0 and X1 = 0, X = 1 So the probability is = 0.5*0.5*3 = 0.75
Question 40
In how many ways can we distribute 5 distinct balls, B1,B2,...,B5 in 5 distinct cells, C1,C2,...,C5 such that Ball B,is not in cell C,Vi=1,2,...,5 and each cell contains exactly one ball?
A
44
B
96
C
120
D
3125
Probability    GATE-IT-2004    
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Question 40 Explanation: 
Total permutations possible = 5!=120 Possible number of ways in which at least one ball is in cell = 5C1*4! - 5C2*3! + 5C3*2! - 5C4*1! + 1 = 76 ->120-76=44 So answer is A
Question 41
A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is
A
1/36
B
1/6
C
1/4
D
1/3
Probability    Gate IT 2005    
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Question 41 Explanation: 
As number of colors is 3,Possible combinations -> 3! = 6 Probability of Blue  marble: 10/60 Probability of Green  marble: 20/60 Probability of Red marble: 30/60 Probability that no two of the marbles drawn have the same colour is = 6 * (10/60 * 20/60 * 30/60) = 1/6 Therefore B is the Answer
Question 42
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is
A
3
B
4
C
5
D
6
Probability    Gate IT 2005    
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Question 42 Explanation: 
  anil_05_32 anil_05_32_1   This solution is contributed by Anil Saikrishna Devarasetty.
Question 43
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 :   Note : This question was asked as Numerical Answer Type.
A
-1
B
0
C
1
D
0.5
Probability    GATE-CS-2016 (Set 1)    
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Question 43 Explanation: 
[Tex] \int_{a}^{1} (\frac{1}{x^{2}}) = \left [ \frac{-1}{x} \right ]_{a}^{1} = \left [ -1 + \frac{1}{a} \right ] [/Tex] But, this is equal to 1. So, (-1) + (1/a) = 1 Therefore, a = 0.5 Thus, D is the correct option.
Question 44
Consider the following experiment.
Step 1. Flip a fair coin twice.
Step 2. If the outcomes are (TAILS, HEADS) then output Y and stop.
Step 3. If the outcomes are either (HEADS, HEAD) or (HEADS, TAILS), 
        then output N and stop.
Step 4. If the outcomes are (TAILS, TAILS), then go to Step 1.
The probability that the output of the experiment is Y is (up to two decimal places). [This Question was originally a Fill-in-the-Blanks question]
A
0.33
B
0.25
C
0.5
D
0.27
Probability    GATE-CS-2016 (Set 1)    
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Question 44 Explanation: 
  A fair coin is flipped twice => {(HH),(HT),(TH),(TT)} four outcomes. Given, if (TH) comes, output ‘Y’. If (HH),or (HT) comes, output ‘N’. If (TT) comes, again flip coin twice. Let, P = probability of getting (TH) or output ‘Y’=1/4. Let, Q= probability of getting (TT) or again flipping coin twice =1/4. Then, probability of getting output ‘Y’= probability of event ‘P’ occurring 1st time (OR) probability of event ‘P’ occurring 2nd time (OR) probability of event ‘P’ occurring 3rd time+……… =P+QP+QQP+QQQP+QQQQP……. =(1/4)+(1/4*1/4)+ (1/4*1/4*1/4)+( 1/4*1/4*1/4*1/4)+…….. It is and infinite GP where, ‘a’=1/4 and ‘r’= 1/4. So, answer= sum= (1/4)/(1-(1/4))=1/3=0.33   This solution is contributed by Sandeep pandey.
Question 45
Suppose that a shop has an equal number of LED bulbs of two different types. 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. The probability that an LED bulb chosen uniformly at random lasts more than 100 hours is :   Note : This question was asked as Numerical Answer Type.
A
0.55
B
0.7
C
0.4
D
0.35
Probability    GATE-CS-2016 (Set 2)    
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Question 45 Explanation: 
The question is based on Bayes' Theorem. P(LED is Type 1) = 1/2 P(LED is type 2) = 1/2 Now, we need to see conditional probabilities. P( LED lasting more than 100 hours / LED is Type 1) = 0.7 P( LED lasting more than 100 hours / LED is Type 2) = 0.4 P(LED lasts more than 100 hours) = P( LED is Type1)* P(LED lasting more than 100 hours / LED is Type 1) + P(LED is Type 2) * P( lasting more than 100 hours / Type 2) = 0.5 * 0.7 + 0.5 * 0.4 = 0.35 + 0.20 = 0.55   Thus, A is the correct choice.
Question 46
Given a coin which gives HEADS with probability 1/4 and TAILS with 3/4. The coin is tossed k times. What is the probability that we get at least k/2 HEADS is less than or equal to?
A
(1/2) k/5
B
(1/2) k/2
C
(1/3) k/2
D
(1/5) k/2
Probability    
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Question 47
When a coin is tossed, the probability of getting a Head is p,0<p<1. Let N be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is
A
1/p
B
1/(1−p)
C
1/p2
D
1/(1−p2)
Misc    Probability    GATE IT 2006    
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Question 47 Explanation: 
  For a continuous variable X ranging over all the real numbers, the expectation is defined by E(X)= ∫ xf(x) dx probability for head = p probability for tail = 1-p if in first time, head appears, probability will be 1*p if firstly tail occurs,and then head occurs, then the probability will be (1-p)*p and so on.... for the nth time, probaility will be (1-p) n-1 * p E= 1*p + 2*(1−p)*p + 3*(1−p)*(1−p)*p + ................... equation(1) multiply both side with (1−p): E*(1-p) = 1*p*(1-p) + 2*(1-p)*(1-p)*p + 3*(1-p)*(1-p)*(1-p)*p +............. equation (2) Subtract equation 2 from equation 1: E−E*(1−p)= 1*p+ (1−p)*p+ (1−p)*(1−p)*p +... E*p =p[1+ (1-p) + (1-p)*(1-p) + ......] It's a infinite geometric progression. E = 1/(1-(1-p)) = 1/p E=1/p correct answer is A. This solution is contributed by Nitika Bansal.
Question 48
A sample space has two events A and B such that probabilities P(A ∩ B) = 1/2, P(A') = 1/3, P(B') = 1/3. What is P(A U B)?
A
11/12
B
10/12
C
9/12
D
8/12
Probability    Gate IT 2008    
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Question 49
What is the probability that in a randomly chosen group of r people at least three people have the same birthday?
2008_23
A
A
B
B
C
C
D
D
Probability    Gate IT 2008    
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Question 49 Explanation: 
anil_22 This solution is contributed by Anil Saikrishna Devarasetty
Question 50
In how many ways can b blue balls and r red balls be distributed in n distinct boxes?
A
[(n+b-1)!(n+r-1)!]/[(n-1)!b!(n-1)!r!]
B
[(n+(b+r)-1)!]/[(n-1)!(n-1)!(b+r)!]
C
n!/(b!r!)
D
[(n+(b+r)-1)!]/[n!(b+r-1)!]
Probability    Gate IT 2008    
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Question 50 Explanation: 
  You have to distribute k balls x 1 ,x 2..... x k that correspond to the number of balls that will be placed in boxes 1, 2, 3 .....k respectively. Since, they should add up to n, you want to determine the number of solutions to the equation: x 1 +x 2 +x 3 +x 4 +......+x k =n solution for the given equation is : (n+k-1)C k = (n+k-1)!/k!(n-1)! For distributing the b blue balls into n distinct boxes, equation will be  x 1 +x 2 +x 3 +....x b = n and solution of the equation will be (n+b-1)C b = (n+b-1)!/b!(n-1)! For distributing the r red balls into n distinct boxes, equation will be x 1 +x 2 +x 3 +....x r = n and solution of the equation will be (n+r-1)C r = (n+r-1)!/r!(n-1)! And For every way of distributing the b blue balls, there are r ways to distribute the red balls, so our total is br. i.e. (n+b-1)! (n+r-1)! / b!(n-1)!r!(n-1)!  which is answer A. Reference: Wikipedia: Stars_and_bars_combinatorics Related: http://www.careerbless.com/aptitude/qa/permutations_combinations_imp8.php This solution is contributed by Nitika Bansal.
Question 51
mock_2
A
A
B
B
C
C
D
D
Probability    GATE 2017 Mock    
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Question 51 Explanation: 
mock_20
Question 52
A random variable x takes values 0, 1, 2, ... with probability proportional to (x+1)(1/5)x. The probability that x <= 5 is:
A
0.6997
B
0.7997
C
0.8997
D
0.9997
Probability    GATE 2017 Mock    
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Question 52 Explanation: 
mock_65
mock_65_1
There are 52 questions to complete.

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