## Probability

 Question 1

Let A and B be any two arbitrary events, then, which one of the following is true?

 A P(A∩B) = P(A)P(B) B P(A∪B) = P(A) + P(B) C P(A|B) = P(A∩B)P(B) D P(A∪B) ≤ P(A) + P(B)
Question 1 Explanation:
(A) Happens when A and B are independent.
(B) Happens when A and B are mutually exclusive.
(C) Not happens.
(D) P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).
 Question 2

The probability of an event B is P1. The probability that events A and B occur together is P2 while the probability that A and occur together is P3. The probability of the event A in terms of P1, P2 and P3 is __________.

 A P2 + P3
Question 2 Explanation:
P(A∩B') = P(A) - P(A∩B)
P3 = P(A) - P2
P(A) = P2 + P3
 Question 3

Let A, B and C be independent events which occur with probabilities 0.8, 0.5 and 0.3 respectively. The probability of occurrence of at least one of the event is __________

 A 0.93
Question 3 Explanation:
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) + P(A∩B∩C)
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)P(B) - P(B)P(C) - P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 - 0.4 - 0.5 - 0.24 + 0.12
= 0.93
 Question 4

The probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7 is:

 A 16/25 B (9/10)3 C 27/75 D 18/25
Question 4 Explanation:
 Question 5

A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession. The probability that one of them is black and the other is white is:

 A 2/3 B 4/5 C 1/2 D 2/1
Question 5 Explanation:
Probability of first ball white and second one black is,

Probability of first ball black and second one white is,
 Question 6

For n>2, let a ∈ {0,1}n be a non-zero vector. Suppose that x is chosen uniformly at random from {0,1}n. Then, the probability that  is an odd number is _____.

 A 0.5
Question 6 Explanation:
‘a’ is a non-zero vector such that a∈{0,1}n
‘x’ is a vector chosen randomly from {0,1}n
‘a’ can have 2(n-1) possibilities, x can have 2n possibilities.
∑aixi have (2n-1)(2n) possibilities, which is an even number of outcomes.
For example:
Take n=3
a = {001, 010, 100, 011, 101, 111}
x = {000, 001, 010, 011, 100, 101, 110, 111}
Computed as [001]×[000] = 0+0+0 = 0 Output = even
[001]×[001] = 0+0+1 = 0 Output = odd
Similarly, there could be 28 even, 28 odd outputs for the a(size=7), x(size=8) of total 56 outputs.
 Question 7

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
Question 7 Explanation:
Cumulative probability is sum of probabilities of all other points till the given value.
The sum of probabilities at x=1, x=-1 itself is 0.5+0.5 = 1. It is evident that, there is no probability for any other values.
The F(x=-1) is 0.5 as per given probabilities and
F(x=1) = sum of F(x=-1) +F(x=0) = ...f(X=1) = 0.5 +0.5 = 1
 Question 8

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
Question 8 Explanation:
The value on the die for first time rolling = {1, 2, 3}
The value on second time can be {1, 2, 3, 4, 5, 6}
So the Sum can be

We have Sample space = 36
The no. of events where (Sum = atleast 6) = {6, 7, 6, 7, 8, 6, 7, 8, 9}
So the probability atleast ‘6’ while getting {1, 2, 3} in first time = 9/36 → ①
If we get ‘6’ in the first time itself, then we do not go for rolling die again.
So, its probability = 1/6
Total probability = 1/6 + 9/36 = 1/6 + 1/4 = 10/24 = 5/12
 Question 9

Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is

 A 1/36 B 1/3 C 25/36 D 11/36
Question 9 Explanation:
1 - no. 6 on both dice
1 - (5/6 × 5/6) = 1 - (25/36) = 11/36
 Question 10

The probability that top and bottom cards of a randomly shuffled deck are both aces in

 A 4/52×4/52 B 4/52×3/52 C 4/52×3/51 D 4/52×4/51
Question 10 Explanation:
E1 : First card being ace
E2 : Last card being ace
Note that E1 and E2 are dependent events, i.e., probability of last card being ace if first is ace will be lesser than the probability of last card being ace if first card is not ace.
So, probability of first card being ace = 4/52
Probability of last card being ace given that first card is ace is,
P(E2 / E1) = 3/51
∴ P(E1 and E2) = P(E1) ⋅ P(E2 / E1) = 4/52 × 3/51
 Question 11

The probability that it will rain today is 0.5. The probability that it will rain tomorrow is 0.6. The probability that it will rain either today or tomorrow is 0.7. What is the probability that it will rain today and tomorrow?

 A 0.3 B 0.25 C 0.35 D 0.4
Question 11 Explanation:
P(Today) = 0.5
P(Tomorrow) = 0.6
P(T∪To) = 0.7
P(T∩To) = P(T) + P(To) - P(T∪To)
= 0.5 + 0.6 - 0.07
= 1.1 - 0.7
= 0.4
 Question 12

A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is

 A 1/6 B 3/8 C 1/8 D 1/2
Question 12 Explanation:
The possibilities of getting exactly one odd number is {OEE, EOE, EEO}.
Total no. of possibilities are 8.
Probability of getting exactly one odd = 3/8
 Question 13
Suppose that the expectation of a random variable X is 5. Which of the following statements is true?
 A There is a sample point at which X has the value 5. B There is a sample point at which X has value greater than 5. C There is a sample point at which X has a value greater than or equal to 5. D None of the above
Question 13 Explanation:
Expectation of discrete random variable
E(X) = x1P1 + x2P2 + ... + xnPn
In question, E(X) is given as 5.
E(X) = 5, 0≤Pi≤1
P1 + P2 + ... + Pn = 1 [Probability]
Therefore, E(X) = 5 is possible only if atleast one of the xi value is greater than 5.
 Question 14

Consider two events E1 and E2 such that probability of E1, Pr[E1]=1/2, probability of E2, Pr[E2]=1/3, and probability of E1 and E2, Pr[E1 and E2]=1/5. Which of the following statement is/are True?

 A Pr[E1 or E2] is 2/3 B Events E1 and E2 are independent C Events E1 and E2 are not independent D Pr[E1/E2] = 4/5
Question 14 Explanation:
If Events E1 and E2 are independent
then P(E1 and E2) = P(E1) × P(E2)
But in the given equation,
P(E1 and E2) = 1/5
P(E1) × P(E2) = 1/2 × 1/3 = 1/6
So, clearly Events E1 and E2 are not independent.
 Question 15

E1 and E2 are events in a probability space satisfying the following constraints:

• Pr(E1) = Pr(E2)
• Pr(E1 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
Question 15 Explanation:
Pr(E1) = Pr(E2) = a(say)
Pr(E1 ∪ E2) = 1
E1 and E2 are Independent.
P(E1 ∪ E2) = Pr(E1) + Pr(E2) - Pr(E1 ∩ E2)
1 = a + a - Pr(E1)⋅Pr(E2)
1 = a + a - a⋅a
2a - a2 = 1
a2 - 2a + 1 = 0
(a - 1)2 = 0
a = 1
Pr(E1) = 1
 Question 16

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
Question 16 Explanation:
Probability of all accidents on sunday = 1/77
Such as total no. of days in a week = 7
Total probability = 7 × 1/77 = 1/76
 Question 17

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
Question 17 Explanation:
Total possibilities = 42 = 16
Atleast one tail = 1/16
Probability of getting one head and one tail is = 1 - 1/16 - 1/16 = 16 - 1 - 1/16 = 14/16 = 7/8
 Question 18

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, 1/4 C 1/2, 1 D 1, 1/2
Question 18 Explanation:
P(A)=1, P(B)=1/2
P(A/B) = P(A∩B)/P(B)
= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)
= 1
P(B/A) = P(B∩A)/P(A)
= P(B)⋅P(A)/P(A)
= 1/2
 Question 19

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 C D max{f1(t), f2(t)}
Question 19 Explanation:
f1(t) and f2(t) are executed sequentially.
→ They representing the probability density functions of time taken to execute.
→ f1 can be executed in 'x' time.
f2 can be executed in 't-x' time.
→ The probability density function =
 Question 20

Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z = min(X,Y), then the mean of Z is given by

 A 1/α+β B min(α, β) C αβ/α + β D α + β
 Question 21

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 Ci, ∀i = 1, 2, …, 5 and each cell contains exactly one ball?

 A 44 B 96 C 120 D 3125
Question 21 Explanation:
Just apply inclusion exclusion principle,
∠5(1 - 1/∠1 + 1/∠2 - 1/∠3 + 1/∠4 - 1/∠5)
= 44
 Question 22

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
Question 22 Explanation:
No. of colours = 3(b,g,r)
Total possible combinations = 3! = 6
Probability of blue marble = 10/60[10 + 20 + 30 = 60]
Probability of green marble = 20/60
Probability of red marble = 30/60
The probability that no two of the marbles has same colour = [10/60 * 20/60 * 30/60] = 1/6
 Question 23

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
Question 23 Explanation:
Generally which uses a recursion strategy to solve this problem. Other than using recursion we can solve using a formula which we know
The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)
The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)
The expected number of coin flips for getting n consecutive same tosses is (2^(n+1) -2) / 2.
where n = 2,
which is (2^(3+1) - 2) / 2 = 3
 Question 24

In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?

 A 0.4 B 0.6 C 0.8 D 0.9
Question 24 Explanation:
Probability rain in afternoon = (0.5×probability of rain when temp≤25) + (0.5×Probability of rain when temp>25)
0.6 = (0.5×0.4) + (0.5×P(rain at temp>25)
0.6 = (2) + (0.5×P(rain at temp>25)
P(rain at temp>25) = 0.8
 Question 25

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)
Question 25 Explanation:
E = 1 × P + (2 × (1 - p) p) + (3 × ( 1 - p) (1 - p) p) + ......
Multiply both sides with (1 - p) and subtract,
E - (1 - p) E = 1 × p + (1 - p) p + (1 - p) (1 - p) p + ......
E - (1 - p) E = p/(1 - (1 - p))
(1 - 1 + p) E = 1
pE = 1
E = 1/p
 Question 26

Suppose there are two coins. The first coin gives heads with probability 5/8 when tossed, while the second coin gives heads with probability 1/4. One of the two coins is picked up at random with equal probability and tossed. What is the probability of obtaining heads ?

 A 7/8 B 1/2 C 7/16 D 5/32
Question 26 Explanation:
Each coin has equal probability to pick i.e., 1/2.
= (1/2)(5/8) + (1/2)(1/4)
= (5/16) + (1/8)
= 7/16
 Question 27

Consider a hash function that distributes keys uniformly. The hash table size is 20. After hashing of how many keys will the probability that any new key hashed collides with an existing one exceed 0.5.

 A 5 B 6 C 7 D 10
Question 27 Explanation:
Total spaces = 20, no. of entries = 1
Probability of collision for each entry = 1/20
After inserting X values then probability becomes 1/2
i.e., (1/20)X = 1/2
X = b
 Question 28

In a multi-user operating system on an average, 20 requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three or five requests are made in 45 minutes is given by :

 A 6.9 × 106 × e-20 B 1.02 × 106 × e-20 C 6.9 × 103 × e-20 D 1.02 × 103 × e-20
Question 28 Explanation:
q0 request in 1 hour. So we can expect 15 requests in 45 minutes.
So, λ=15
 Question 29

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 ∪ B)?

 A 11/12 B 10/12 C 9/12 D 8/12
Question 29 Explanation:
P(A ∩ B) = 1/2
P(A') = 1/3; P(A) = 2/3
P(B') = 1/3; P(B) = 2/3
P(A ∪ B) = P(A) +P(B) - P(A ∩ B)
= 2/3 + 2/3 - 1/2
= 4+4-3/ 6
= 5/6
= 10/12
 Question 30

What is the probability that in a randomly chosen group of r people at least three people have the same birthday?

 A B C D
Question 30 Explanation:
 Question 31

Which of the following statements are FALSE?

 A For poisson distribution, the mean is twice the variance. B In queuing theory, if arrivals occur according to poisson distribution, then the inter-arrival time is exponentially distributed. C The distribution of waiting time is independent of the service discipline used in selecting the waiting customers for service. D If the time between successive arrivals is exponential, then the time between the occurrences of every third arrival is also exponential. E Both (A) and (C).
Question 31 Explanation:
In Poisson distribution, the mean is not equal to the twice the variance.
Option C is also False, because waiting time is dependent on the service discipline.
 Question 32

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 3/4
Question 32 Explanation:
A fair coin is tossed 4 times.
Then total number of possibilities = 24 = 16
No. of possibilities getting 2 heads and 2 tails is
HHTT, HTHT, TTHH, THTH, THHT, HTTH = 6
Probability of getting 2 heads and 2 tails is
= No. of possibilities/Total no. of possibilities = 6/16 = 3/8
 Question 33

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 probability. The sum total of the expected marks obtained by all these students is:

 A 0 B 2550 C 7525 D 9375
Question 33 Explanation:
Probability of choosing a correct answer = 1/4
Probability of selecting a wrong answer = 3/4
Expected marks for each question = (1/4) × 1 + (3/4) -(0.25)
= 1/4 + (-3/16)
= 4-3/16
= 1/16
= 0.0625
Expected marks for 150 questions = 150 × 0.625 = 9.375
The sum total of expected marks obtained by 1000 students is = 1000×9.375 = 9375
 Question 34

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 / 2d C d/2n D 1/2d
Question 34 Explanation:
n binary bits with difference 'd' then no. of favourable cases = nCd
Total no. of cases where n positions have any binary bit = 2n
The probability of 'd' bits differ = nCd / 2n
 Question 35

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
Question 35 Explanation:

Above diagram shows the scenario of our question.
The length p of our position vector (x,y) is
p = √x2 + y2
p2 = x2 + y2
E(p2) = E(x2 + y2) = E(x2) + E(y2)
Now we need to calculate the probability density function of X and Y.
Since distribution is uniform,
X goes from 0 to 1, so PDF(x) = 1/1-0 = 1
Y goes from 0 to 2, so PDF(y) = 1/2-0 = 1/2
Now we evaluate,

E(p2) = E(x2) + E(y2) = 5/3
 Question 36

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 __________.

 A 0.25 B 0.26 C 0.27 D 0.28
Question 36 Explanation:
We know that
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2 ≥ √(P(A)∙P(B))
1/2 ≥ √(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4 ≥ P(A)∙P(B)
P(A)∙P(B) ≤ 1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
There are 36 questions to complete.

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