Probability

  • Last Updated : 21 Jan, 2014

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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)
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 \lambda = 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    50 Aptitude & Reasoning MCQs with Answers    
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Question 2 Explanation: 
There are total 90 two digit numbers, out of them 13 are divisible by 7, these are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98. Therefore, probability that selected number is not divisible by 7 = 1 - 13/90 = 77/90. So, option (D) is true.
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: 

Solution set: { 6, (1,5), (1,6) ......}

                  i.e. P(6 appeared on first throw) +

                       P(1 appeared on first throw and 5 appeared on second throw) +

                        P(1 appeared on first throw and 6 appeared on second throw) + ....................

                      = 1/6 + (1/6)(1/6) + (1/6)(1/6) + .....

                      = 1/6 + 9/36

                      = 5/12.

Viewpoint 2:

P(......) =   P(6 came on first throw) +  P(sum>= 6 and 1,2,3 appeared in first throw)

          =      1/6                                +                 ????

P(1,2,3 appeared in first throw) = 1/2                                                   //P(E1)

P(sum >= 6 | 1,2,3 appeared in first throw) = 9/18                             //P(E2 | E1)

// Our new sample space is:  { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

                                                 (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

                                                (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) }

// 9 favorable cases: {(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6) }

P(sum>= 6 and 1,2,3 appeared in first throw) = (1/2)(9/18)              //P(E2 ∩ E1) = P(E1)P(E2|E1)

P(what we are looking for) = 1/6 + 9/36     = 5/12

Correct Answer: B

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    50 Aptitude & Reasoning MCQs with Answers    
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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 Y_{i} = a X_{i} + b \newline \newline \sum Y_{i} = \sum(aX-{i} + b) \newline \sum Y_{i} = a(\sum X_{i}) + nb \newline (\sum Y_{i})/n = a(\sum X_{i})/n + b \newline \mu _{y} = a\mu_{x} + b And for the standard deviation \newline \newline \sigma _{y} = \sqrt{\frac{1}{n} \sum (\mu_{y} - Y_{i})^{2}} \newline \newline \sigma _{y} = \sqrt{\frac{1}{n} \sum (a\mu_{x} + b - Y_{i})^{2}} \newline \sigma _{y} = \sqrt{\frac{1}{n} \sum (a\mu_{x} + b - aX_{i} - b)^{2}} \newline \newline \sigma _{y} = \sqrt{\frac{1}{n} \sum (a\mu_{x} - aX_{i} )^{2}} \newline \newline \sigma _{y} = a\sqrt{\frac{1}{n} \sum (\mu_{x} - X_{i} )^{2}} \newline \newline \sigma _{y} = a\sigma_{x}    
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    50 Aptitude & Reasoning MCQs with Answers    
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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, 100x1096, 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
There are 94 questions to complete.
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