Probability
Question 1 
1/5  
1/3  
1/4  
2/5 
Total number of seats = 5*40
= 40 seats
Total number of students= 30
Step2: Given constraint that, 6th seat in the fifth row is empty
When we are deleting 6th seat, 30 students have 39 choices of seats
Step3: Total number of choices = ^{39}C_{30}
Total ways to choose = ^{40}C_{30}
Step4: Final probability = ^{39}C_{30} /^{ 40}C_{30}
= 1/4
Question 2 
2/9  
4/9  
2/3  
1/2 
Probability that chip came chosen from company Y = 4/9
Number of Defective chips from company X = 1
Number of Defective chips from company Y = 2
Probability that chip is defective from company X = 5/9 * 1/5
Probability that chip is defective from company Y = 4/9 * 2/4
Probability that chip is defective = 5/9 * 1/5 + 4/9 * 2/4
Given chip is defective, probability that it came from the company
Y = P(Defective Company Y)/ P(Defective)
= (4/9 * 2/4) / (5/9 * 1/5 + 4/9 * 2/4)
= 2/3
Question 3 
f_{1}(t) + f_{2}(t)  
∫^{t}_{0} f_{1}(x), f_{2}(x) dx  
∫^{t}_{0} f_{1}(x), f_{2}(tx) dx  
max (f_{1}(t), f_{2}(t)) 
∫^{t}_{0} f_{1}(x), f_{2}(tx) dx
Question 4 
P(A∩B) = P(A)P(B)  
P(A∪B) = P(A) + P(B)  
P(AB) = P(A ∩ B) + P(B)  
P(A∪B) <= P(A) + P(B) 
(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 5 
1  
1/21  
0  
0.5 
Step1: Bag contains 19 Red(R) and 19 Blue(B) balls.
BB (or) RR happen we are discarded.
If we get BR (or) RB then B is discarded and R is returned.
Step2: There are some conditions that,
→ If black balls will either come with black then both black balls are discarder.
→ If it will come with red then only black balls will be discarded.
→ Suppose 2 red balls will come together means we are discarding both red balls.
Step3: As per the above constraints, total 19 Red balls it means odd number.
→ Among 19 only 18 will be discarded.
Step3: Final content of bag at second last trail will be either R,B (or) R,R,R and finally in last
trail bag will left with one red ball in both the cases.
Question 6 
f (b − a)  
f (b) − f (a)  
Question 7 
5/8  
1/8  
2/3  
3/8 
The combinations are (HHH,HHT,HTH,HTT,TTT,TTH,THT,THH)
The number of combinations with two heads and one tail is HHT,HTH,THH
The the probability is the number of combinations of the event/ total combinations of the event = 3/8
Question 8 
0.5, 0.25  
0.25, 0.5  
0.5, 1  
1, 0.5 
→P(E) denote the probability of the occurrence of event E
→P(A) = 0.5 and P(B) = 1
→Conditional probability is a measure of the probability of an event (some particular situation occurring) given that another event has occurred.
→If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A  B), or P_{B}(A)
→P(A/B) = P(A ∩ B)/P(B)
→If two events A and B are independent, then the probability of both events is the product of the probabilities for each event: P(A ∩ B) = P(A)P(B)
→P(A/B) = P(A) * P(B) / P(B)
→P(A/B) = 0.5
→Similarly, P(B/A) = P(A) * P(B) / P(A) and P(B/A) = 1
Question 9 
30  
25  
22.5  
27.5 
Mode = 3 Median – 2 Mean
15 = 3 Median  2(30)
Median = 25
Question 10 
1/6  
1/12  
1/144  
1/24 
Probability of both friends born in a same month is (1/12) * (1/12)
Suppose, they born in january = 1/12 * 1/12
Suppose, they born in february = 1/12 * 1/12
; ; Suppose, they born in December = 1/12 * 1/12
Probability that two friends are born in the same month is= 12*(1/12)*(1/12)
= 1/12
Question 11 
Consider the set of all possible fivecard poker hands dealt fairly from a standard deck of fiftytwo cards. How many atomic events are there in the joint probability distribution ?
2,598,960
 
3,468,960  
3,958,590
 
2,645,590

Given random variables X, Y, …, that are defined on a probability space, the joint probability distribution for X, Y, … is a probability distribution that gives the probability that each of X, Y, … falls in any particular range or discrete set of values specified for that variable. In the case of only two random variables, this is called a bivariate distribution, but the concept generalizes to any number of random variables, giving a multivariate distribution.
Given data,
 Total cards = 52
 poker hand = 5 cards of all possibilities
Step1: This problem, we are simply finding combinations of ^{52}C_{5}
= C(n,r) = C(52,5)
= 52! / (5!(525)!)
= 2598960
Question 12 
Digital data received from a sensor can fill up 0 to 32 buffers. Let the sample space be S = {0, 1, 2, .........., 32} where the sample j denote that j of the buffers are full and P(i) = 1/561 (33i).
Let A denote the event that the even number of buffers are full. Then p(A) is:0.515  
0.785  
0.758  
0.485 
• Probability of all even number of buffers are full is P(A).
• We are going find the values of P(0),P(2),P(4),P(6) …. P(16).
• P(0) is 1/562 (330)
• P(2) is 1/562 (332)
• …
• P(16) is 1/562 (3332)
• The probability of all even number of buffers are full is P(A) which is equal to P(0)+P(2)+P(4)+P(6)+ …. P(16).
• P(A) = 1/562(33+31+29+27+....+1)
• P(A) = 1562×289 = 0.51423
Question 13 
The number of substrings that can be formed from string given by
“a d e f b g h n m p” is
10  
45  
56  
55 
n*(n+1)/2 + 1
We have added 1 because it may include a NULL string also.
The number of substrings = 10*(11)/2 + 1
The number of substrings = 56
Question 14 
A survey has been conducted on methods of commuter travel. Each respondent was asked to check Bus, Train or Automobile as a major methods of travelling to work. More than one answer was permitted. The results reported were as follows :
Bus 30 people; Train 35 people; Automobile 100 people; Bus and Train 15 people; Bus and Automobile 15 people; Train and Automobile 20 people; and all the three methods 5 people. How many people complete the survey form ?
160  
120  
165  
115 
Question 15 
A full joint distribution for the Toothache, Cavity and Catch is given in the table below:
Which is the probability of Cavity, given evidence of Toothache ?
< 0.2, 0.8 >
 
< 0.6, 0.8 >
 
< 0.4, 0.8 >
 
< 0.6, 0.4 >

Question 16 
160  
120  
165  
115 
Let “B” denotes people travelled by Train.
Let “C” denotes people travelled by Automobile.
Let “(A⋂B)” denotes people travelled by Bus and Train.
Let “(A⋂C)” denotes people travelled by Bus and Automobile.
Let “(B⋂C)” denotes people travelled by Train and Automobile.
Let “(A⋂B⋂C)” denotes people travelled by Bus, Train and Automobile.
Number of people completed the survey = A+B+C(A⋂B)(A⋂C)(B⋂C)+(A⋂B⋂C)
= 30+35+100151520+5
= 120
Question 17 
Which is the probability of Cavity, given evidence of Toothache ?
< 0.2, 0.8 >  
< 0.6, 0.8 >  
< 0.4, 0.8 >  
< 0.6, 0.4 > 
P(Cavity/Toothache)=P(Cavity ⋀ Toothache) / P(toothache)
=(0.108 + 0.012) / (0.108 + 0.012 + 0.016 + 0.064)
=(0.12)/(0.2)
=0.6
P(~Cavity / Toothache) = (0.016 + 0.064) / 0.2
= 0.4
Question 18 
1/35  
1/14  
1/9  
3/7 
→ Green balls are 4
→ Total 10 balls in the box. We need to select 4 balls from 10 balls in which two red balls out of 6 red balls and two green balls from 4 green balls.
→ Total of number of ways for selecting 4 balls out of 10 balls is C(10,4)
→ Number of ways for selecting two red balls from 6 and two green balls from 4 is C(6,2)*C(4,2)
→ Probability of selecting two balls is number of ways for selecting four balls / total number of ways P= C(6,2)*C(4,2) / C(10,4) = C(6,2) is 6x5/2=15
= C(4,2) is 4x3/2=6
= C(10,4) is (10x9x8x7) / (4x3x2x1)=210
P=C(6,2)*C(4,2) / C(10,4)
=15x6/210=
= 3/7, So option 4 is correct
Question 19 
4/52 X 4/52  
4/52 X 3/52  
4/52 X 3/51  
4/52 X 4/51 
E 2 : Last card being ace
Note that E 1 and E 2 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(E 2 / E 1 ) = 3/51
∴ P(E 1 and E 2 ) = P(E 1 ) ⋅ P(E 2 / E 1 ) = 4/52 * 3/51
Question 20 
Which one of the following is most affected by the presence of outliers in sample data?
Variance
 
Mean
 
Median
 
Mode

For the sample data set:
1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4
We find the following mean, median, mode, and standard deviation:
Mean = 2.58
Median = 2.5
Mode = 2
Standard Deviation = 1.08
If we add an outlier to the data set:
1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 400
The new values of our statistics are:
Mean = 35.38
Median = 2.5
Mode = 2
Standard Deviation = 114.74
Note: Outliers often has a significant effect on your mean and standard deviation.
Question 21 
A fair coin is tossed 6 times. What is the probability that exactly two heads will occur?
11/32  
1/64  
15/64
 
63/64 
The probability can be calculated as:
n is a total number of experiments
k is an expected number of successes
p is the probability of a success
Step2: The first task can be simply solved by calculating the probability of 2 successes:
A "success" is "tossing heads in a single toss".
A "failure" is "tossing tails in a single toss".
Step3: The probability of a success is ½
The number of all experiments is n=6.
The number of successes is k=2
Question 22 
If the mean of a poisson distribution is m, then standard deviation of the distribution is:
m^{2}  
m  
2*m  
√m 
Question 23 
The standard deviation of binomial distribution with n observations and probability of success p, probability of failure is q is:
√npq
 
Pq  
Np  
√pq 
The binomial distribution for a random variable X with parameters n and p represents the sum of n independent variables Z which may assume the values 0 or 1. If the probability that each Z variable assumes the value 1 is equal to p, then the mean of each variable is equal to l*p + 0*(lp) = p, and the variance is equal to p(lp).
By the addition properties for independent random variables, the mean and variance of the binomial distribution are equal to the sum of the means and variances of the n independent Z variables, so
These definitions are intuitively logical. Imagine, for example 8 flips of a coin. If the coin is fair, then p = 0.5. One would expect the mean number of heads to be half the flips, or np = 8*0.5 = 4.
The variance is equal to np(lp) = 8*0.5*0.5 = 2.
Question 24 
The normal curve is symmetrical about its:
Standard deviation  
Mean
 
Variance  
Probability

The probability density of the normal distribution is
Where
• μ is the mean or expectation of the distribution (and also its median and mode).
• σ is the standard deviation, and
• 2 is the variance
Question 25 
The density of uniform distribution over the interval ⍺ < a < b < ⍺ is given by:
f(x) = λe^{λx} , x>=0
 
f(x) = q^{k}p  
f(x) = 1/(ba), a  
f(x) = (⍺/c)x^{⍺1}

The values of f(x) at the two boundaries a and b are usually unimportant because they do not alter the values of the integrals of f(X) dX over any interval, not of X f(X) dX or any higher moment. Sometimes they are chosen to be zero, and sometimes chosen to be 1/(b – a). The latter is appropriate in the context of estimation by the method of maximum likelihood. In the context of Fourier analysis, one may lake the value of f(a) to be 12(b – a), since then the inverse transform of may integral transform of this uniform function will yield back the function itself, rather than a function which equal ‘almost everywhere’, i.e except on a set of points with zero measure. Also, it is consistent with the sign function which has no such ambiguity.
In terms of mean and variance σ^{2}, the probability density may be written as:
Question 26 
Exponential distribution is special case of ____ distribution.
Theta  
Alpha  
Beta  
Gamma 
Question 27 
0  
1  
1  
⍺ 
Question 28 
The mean, mode and median are connected by the empirical relationship:
Meanmode = 2(meanmedian)
 
Meanmode = 3(meanmedian)  
Meanmode = (meanmode)/2
 
Meanmode = (meanmode)/3

Mode = mean  3 [mean  median]
Mode = 3 median  2 mean
and Median = mode + ⅔ [meanmode]
Question 29 
If a random variable takes a finite set of values it is called:
Continuous variate  
Normal variate
 
Discrete variate
 
Exponential variate 
Examples:
number of students present
number of red marbles in a jar
number of heads when flipping three coins
students’ grade level
→ A discrete random variable X has a countable number of possible values.
Example:
Let X represent the sum of two dice.
→ A discrete random variable is one which may take on only a countable number of distinct values such as 0,1,2,3,4,........ Discrete random variables are usually (but not necessarily) counts. If a random variable can take only a finite number of distinct values, then it must be discrete.
Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's surgery, the number of defective light bulbs in a box of ten.
Question 30 
The root mean square deviation when measured from the mean is:
Greatest
 
Positive
 
Least
 
Negative

Question 31 
Portability is not a quality factor of:
Software coding
 
Software design
 
Software Process  
Software testing

This model classifies all software requirements into 11 software quality factors. The 11 factors are grouped into three categories – product operation, product revision, and product transition factors.
1. Product operation factors − Correctness, Reliability, Efficiency, Integrity, Usability.
2. Product revision factors − Maintainability, Flexibility, Testability.
3. Product transition factors − Portability, Reusability, Interoperability.
Question 32 
720  
120  
60  
360 
BANANA= 6 letters
B is appearing 1 time
A is appearing 3 times
N is appearing 2 times
So, = 6! / (3! * 2!)
= 60
Question 33 
1/64  
1/32  
1/8  
1⁄4 
Ex: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121...,
Step1: Here, string length is 10. If we consider first 5 numbers as random choices like 0 or 1.
Remaining 5 numbers are fixed. Total number of possibilities are 2^{ 10} . But we are only considering first 5 choices. The probability is 2 ^{5} .
Step2: The probability 2 ^{5} /2^{ 10}
= 1/2 ^{5}
= 1/32
Question 34 
3/4  
2/3  
1/2  
1⁄3 
 3 cards in a box
 1 st card: Both sides of one card is black. The card having 2 sides. We can write it as BB.
 2 nd card: Both sides of one card is red. The card having 2 sides. We can write it as RR.
 3rd card: one black side and one red side. We can write it as BR.
Step1: The probability that the opposite side is the same colour as the one side we observed is 2⁄3 because total number of cards are 3
Question 35 
75240  
52492  
41800  
9900 
 20 men and 12 women
 5 people can choose from men and women
 Each committee contains at least three women
Step1: We have a constraint that each committee contains atleast 3 women.
possibility1: 2 men + 3 women.
possibility2: 1 men + 4 women.
possibility3: 0 men + 5 women.
Step1: They are asking to find all possibilities.
= (possibility 1)+ (possibility 2) + (possibility 3)
= . ( ^{20} C_{ 2}. * ^{12} C_{ 3} ) + ( ^{20} C_{ 2}. * ^{12} C _{2} ) + ( ^{20} C_{ 2}. *^{ 12} C_{ 2} )
= (190*220) + (20*495) + (1*792)
= 41800 + 9900 + 792
= 52492
Question 36 
22 / 36  
12 / 36  
14 / 36  
6 / 36 
→ The possible ways are 1,2,3,4,5,6
→ The possible ways of one number divides another number are (1,1) (1,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (5,5) and (6,6).
→ The probability is the number of outcomes / total outcomes = 14/36
Question 37 
P(A∩B) = P(A) P(B)  
P(P∪B) = P(A) + P(B)  
P(A∪B) ≤ P(A) + P(B)  
P(A / B) = P(A∩B) + P(B) 
OptionB is happens when A and B are mutually exclusive.
OptionC is not happens.
OptionD is P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B)  P(A∩B).
Question 38 
533  
599  
467  
66 
Given data,
 Set A={1, 2, 3, ........, 1000}
 Set A shall be divisible by 3=?
 Set A shall be divisible by 5=?
 Set A shall be divisible by 3 and 5=?
Step1: To find divisible by 3 numbers are
=⌊1000/3⌋
= 333
Step2: To find divisible by 5 numbers are
=⌊1000/5⌋
= 200
Step3: To find divisible by 3 and 5 numbers are
=⌊1000/(3*5)⌋
= 66
These 66 is already part of 333 and 200. So, we have to exclude it from the list.
Total= 333+20066
= 467
Note: We are using floor because excluding fraction value.
Method2:
The above problem is in the form of (AUB) = (A)+(B)(A∩B)
A=1000/3
B=1000/5
(A∩B)=66
(AUB)=467
Note: Getting this idea in exam hall is very difficult. So, better follow method1 ratherhan method2
Question 39 
0.5  
0.704  
0.632121  
0.68 
1p(x=0)
We will apply the formula of binomial distribution
1(^{ 1000000} C _{0} (1/1000000) ^{0} * (999999/1000000) ^{10^6} )
=1(1*1*0.367879)
=0.632121
Hence answer is option C
Question 40 
e^{ 15}  
1e ^{15}  
1e ^{20}  
e ^{20} 
→ The time between each pair of consecutive events has an exponential distribution with parameter λ and each of these interarrival times is assumed to be independent of other interarrival times.
→ λ = (40*30)/60
=20
P(T>40min)= 1P(T ≤ 40 min)
=1(1e ^{40 min / λ} )
=1(1e^{ 40 min / 20} )
= e^{ 20}
Question 41 
2,598,960  
3,468,960  
3,958,590  
2,645,590 
Given random variables X,Y, ... , that are defined on a probability space, the joint probability distribution for X,Y, ... is a probability distribution that gives the probability that each of X,Y, ... falls in any particular range or discrete set of values specified for that variable. In the case of only two random variables, this is called a bivariate distribution, but the concept generalizes to any number of random variables, giving a multivariate distribution.
Given data,
Total cards=52
poker hand=5 cards of all possibilities.
Step1: This problem, we are simply finding combinations of ^{52} C _{5}
= C(n,r)=C(52,5)
= 52! / (5!(525)!)
= 2598960
Question 42 
0.515  
0.785  
0.758  
0.485 
Probability of all even number of buffers are full is P(A)
We are going find the values of P(0),P(2),P(4),P(6) .... P(16).
P(0) is 1/562 (330)
P(2) is 1/562 (332)
...
P(16) is 1/562 (3332)
The probability of all even number of buffers are full is P(A) which is equal to
P(0)+P(2)+P(4)+P(6)+ .... P(16).
P(A) = 1/562(33+31+29+27+....+1)
P(A) =1562×289=0.51423
Question 43 
25  
20  
35  
30  
None of the above 
 Hash table has space =75 slots.
 For 6% filling it must take =6 slots.
 Probability of no collision for first 6 entries=?
Step1: According to given data,
= (^{75}P_{6) / (756) = 0.814586387 Step2: We have to find at least one collision occurs in 6 entries =1Probability of no collision for first 6 entries = 10.814586387 = 0.185413613 Note: Given options are wrong. Given optionB is correct answer according to original key. }
Question 44 
e^{–15 }  
e^{–5}  
1 – e^{–5 }  
1 – e^{–10} 
An event can occur 0, 1, 2, … times in an interval. The average number of events in an interval is designated (lambda). is the event rate, also called the rate parameter. The probability of observing k events in an interval is given by the equation
P(k events in interval)=e^{λ} [(λ^{k})/k!]
Where
→ λ is the average number of events per interval
→ e is the number 2.71828... (Euler's number) the base of the natural logarithms
→ k takes values 0, 1, 2, …
→ k! = k × (k − 1) × (k − 2) × … × 2 × 1 is the factorial of k.
→ Now it is given that in 1 hour (60 minutes) 20 requests are made
→ In 1 minute number of requests made= 20/60
→ In 45 minutes total number of requests made = (20/60) * 45
= 15
Probability= (e^{15} *(15)^{0}) / 0!
= e^{15}
Question 45 
963/1000  
973/1000  
983/1000  
953/1000  
None of the above 
Probability of not getting a number = 1 1/(10) = 9/10
Now probability that correct number is chosen in first chance= 1/10
probability that correct number is chosen in second chance= (9/10)* (1/10)
probability that correct number is chosen in first chance= (9/10)* (9/10) *(1/10)
So total probability = (1/10)+ [(9/10)* (1/10) ]+ [(9/10)* (9/10) *(1/10)]
= (1/10)+(9/100)+(81/1000)
= 271/1000
Note: Excluded for evaluation because given options are wrong
Question 46 
e ^{−15}  
e ^{−5}  
1 − e^{ −5}  
1 − e^{ −10} 
Question 47 
p(x, y)=p(x).p(y)  
p(x, y)=p(x)+p(y)  
p(x, y)=p(x+y)  
p(x, y)=p(x).p(y)+p(x) 
Note: Original key given optionC is correct answer
Question 48 
3/7  
4/7  
5/7  
6/7 
Question 49 
nP(1 – P)^{n – 1}  
nP  
P(1 – P)^{n – 1}  
n^{P}(1 – P)^{n – 1} 
Question 50 
2 bits per symbol  
1.75 bits per symbol  
1.50 bits per symbol  
1.25 bits per symbol 
Question 51 
Initially, we have w(i,j)=q(i), c(i,j)=0 and r(i,i)=0, 0 ≤ i ≤ 4. The observations are w(i,j)=p(j)+q(j)+w(i,j1), we get
w(0,1) = p(1) + q(1) + w(0,0) = 8
c(0,1) = w(0,1) + min{c(0,0) + c(1,1)} = 8
r(0,1) = 1
w(1,2) = p(2) + q(2) + w(1,1) = 7
c(1,2) = w(1,2) + min{c(1,1) + c(2,2)} = 7
r(0,2) = 2
w(2,3) = p(3) + q(3) + w(2,2) = 3
c(2,3) = w(2,3) + min{c(2,2) + c(3,3)} = 3
r(2,3) = 3
w(3,4) = p(4) + q(4) + w(3,3) = 3
c(3,4) = w(3,4) + min{c(3,3) + c(4,4)} = 3
r(3,4) = 4 knowing w(i, i+1) and c(i, i+1), 0 ≤ i < 4 and to compute recurrence relation of OBST and compute w(i, i+2), c(i,i+2) and r(i,i+2), 0 ≤ i < 3. This process can repeated until w(0,4), c(0,4) and r(0,4) are obtained.
Question 52 
is where p is the transition probability. Then the conditional entropy is given by:
1  
 plog(p)  (1  p)log(1  p)  
1 + p log(p) + (1  p)log(1  p)  
0 