Probability

Question 1
A class of 30 students occupy a classroom containing 5 rows of seats, with 8 seats in each row. If the students seat themselves at random, the probability that the sixth seat in the fifth row will be empty is
A
1/5
B
1/3
C
1/4
D
2/5
       Engineering-Mathematics       Probability       ISRO-2018
Question 1 Explanation: 
Step-1: Given, 5 rows with 8 seats in each row.
Total number of seats = 5*40
= 40 seats
Total number of students= 30
Step-2: Given constraint that, 6th seat in the fifth row is empty
When we are deleting 6th seat, 30 students have 39 choices of seats
Step-3: Total number of choices = 39C30
Total ways to choose = 40C30
Step-4: Final probability = 39C30 / 40C30
= 1/4
Question 2
Company X shipped 5 computer chips, 1 of which was defective and company Y shipped 4 computer chips, 2 of which were defective. One computer chip is to be chosen uniformly at a random from the 9 chips shipped by the companies. If the chosen chip is found to be defective, what is the probability that the chip came from the company Y?
A
2/9
B
4/9
C
2/3
D
1/2
       Engineering-Mathematics       Probability       ISRO-2007
Question 2 Explanation: 
Probability that chip came chosen from company X = 5/9
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
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
t0 f1(x), f2(x) dx
C
t0 f1(x), f2(t-x) dx
D
max (f1(t), f2(t))
       Engineering-Mathematics       Probability       ISRO-2007
Question 3 Explanation: 
The two modules executed sequentially. The total runtime of the program is the sum of runtime of the two module. Probability density function of overall time taken
t0 f1(x), f2(t-x) dx
Question 4
If A and B be two arbitrary events, then
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)
       Engineering-Mathematics       Probability       ISRO-2017 May
Question 4 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 5
A bag contains 19 red balls and 19 black balls. Two balls are removed at a time repeatedly and discarded if they are of the same colour, but if they are different, black ball is discarded and red ball is returned to the bag. The probability that this process will terminate with one red ball is
A
1
B
1/21
C
0
D
0.5
       Engineering-Mathematics       Probability       ISRO-2017 December
Question 5 Explanation: 
Given data,
Step-1: 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.
Step-2: 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.
Step-3: As per the above constraints, total 19 Red balls it means odd number.
→ Among 19 only 18 will be discarded.
Step-3: 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
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
D
       Engineering-Mathematics       Probability       ISRO CS 2009
Question 6 Explanation: 
A non-discrete random variable X is said to be absolutely continuous, or simply continuous, if its distribution function may be represented as
Question 7
Three coins are tossed simultaneously. The probability that they will fall two heads and one tail is
A
5/8
B
1/8
C
2/3
D
3/8
       Engineering-Mathematics       Probability       ISRO CS 2011
Question 7 Explanation: 
Three coins tossed means , total number of combinations(possibilities) are 23=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
Let P(E) denote the probability of the occurrence of event E. If P(A) = 0.5 and P(B) = 1, then the values of P(A/B) and P(B/A) respectively are
A
0.5, 0.25
B
0.25, 0.5
C
0.5, 1
D
1, 0.5
       Engineering-Mathematics       Probability       ISRO CS 2013
Question 8 Explanation: 
Given data is
→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 PB(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
What is the median of data if its mode is 15 and the mean is 30?
A
30
B
25
C
22.5
D
27.5
       Engineering-Mathematics       Probability       ISRO CS 2014
Question 9 Explanation: 
In a moderately symmetric distribution, the mean, median and mode are connected by the formula:
Mode = 3 Median – 2 Mean
15 = 3 Median - 2(30)
Median = 25
Question 10
The probability that two friends are born in the same month is?
A
1/6
B
1/12
C
1/144
D
1/24
       Engineering-Mathematics       Probability       ISRO CS 2014
Question 10 Explanation: 
Probability of a person to be born in a one month out of 12 months is 1/12
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 five-card poker hands dealt fairly from a standard deck of fifty-two cards. How many atomic events are there in the joint probability distribution ?

A
2,598,960
B
3,468,960
C
3,958,590
D
2,645,590
       Engineering-Mathematics       Probability       UGC-NET JUNE Paper-2
Question 11 Explanation: 
Joint probability distribution:
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
Step-1: This problem, we are simply finding combinations of 52C5
= C(n,r) = C(52,5)
= 52! / (5!(52-5)!)
= 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 (33-i).

Let A denote the event that the even number of buffers are full. Then p(A) is:
A
0.515
B
0.785
C
0.758
D
0.485
       Engineering-Mathematics       Probability       UGC-NET JUNE Paper-2
Question 12 Explanation: 
• Probability of ith buffer getting full = p(i) = 1/562(33−i)
• 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 (33-0)
• P(2) is 1/562 (33-2)
• …
• P(16) is 1/562 (33-32)
• 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

A
10
B
45
C
56
D
55
       Engineering-Mathematics       Probability       UGC-NET DEC Paper-2
Question 13 Explanation: 
If we have no repetition in a string then the number of substrings can be found using the formula :
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 ?

A
160
B
120
C
165
D
115
       Engineering-Mathematics       Probability       UGC-NET DEC Paper-2
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 ?

A
< 0.2, 0.8 >
B
< 0.6, 0.8 >
C
< 0.4, 0.8 >
D
< 0.6, 0.4 >
       Engineering-Mathematics       Probability       UGC-NET DEC Paper-2
Question 16
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 ?
A
160
B
120
C
165
D
115
       Engineering-Mathematics       Probability       UGC NET CS 2018-DEC Paper-2
Question 16 Explanation: 
Let “A” denotes people travelled by Bus.
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+100-15-15-20+5
= 120
Question 17
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 ?
A
< 0.2, 0.8 >
B
< 0.6, 0.8 >
C
< 0.4, 0.8 >
D
< 0.6, 0.4 >
       Engineering-Mathematics       Probability       UGC NET CS 2018-DEC Paper-2
Question 17 Explanation: 
Probability cavity given that toothache
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
A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls will be red and two will be green ?
A
1/35
B
1/14
C
1/9
D
3/7
       Engineering-Mathematics       Probability       UGC NET CS 2018-DEC Paper-2
Question 18 Explanation: 
→ Total red balls are 6
→ 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
The probability that top and bottom cards of a randomly shuffled deck are both aces is:
A
4/52 X 4/52
B
4/52 X 3/52
C
4/52 X 3/51
D
4/52 X 4/51
       Engineering-Mathematics       Probability       Nielit Scientist-B IT 22-07-2017
Question 19 Explanation: 
E​ 1​ : First card being ace
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?

A
Variance
B
Mean
C
Median
D
Mode
       Engineering-Mathematics       Probability       JT(IT) 2018 PART-B Computer Science
Question 20 Explanation: 
Let's examine what can happen to a data set with outliers.
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?

A
11/32
B
1/64
C
15/64
D
63/64
       Engineering-Mathematics       Probability       JT(IT) 2018 PART-B Computer Science
Question 21 Explanation: 
Step-1: We want to calculate the probability of getting exactly k "success" results using Bernoulli's Scheme.

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
Step-2: 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".
Step-3: 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:

A
m2
B
m
C
2*m
D
√m
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 22 Explanation: 
Mean and Variance of the Poisson distribution. There is also a formula for the standard deviation, σ, and variance, σ2.
Question 23

The standard deviation of binomial distribution with n observations and probability of success p, probability of failure is q is:

A
√npq
B
Pq
C
Np
D
√pq
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 23 Explanation: 
Mean and Variance of the Binomial Distribution:
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*(l-p) = p, and the variance is equal to p(l-p).
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(l-p) = 8*0.5*0.5 = 2.
Question 24

The normal curve is symmetrical about its:

A
Standard deviation
B
Mean
C
Variance
D
Probability
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 24 Explanation: 
Symmetrical distribution occurs when the values of variables occur at regular frequencies and the mean, median and mode occur at the same point. In graph form, symmetrical distribution often appears as a bell curve. If a line were drawn dissecting the middle of the graph, it would show two sides that mirror each other.
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:

A
f(x) = λe-λx , x>=0
B
f(x) = qkp
C
f(x) = 1/(b-a), a
D
f(x) = (⍺/c)x⍺-1
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 25 Explanation: 
The probability density function of the continuous uniform distribution is:

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 1|2(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 σ2, the probability density may be written as:
Question 26

Exponential distribution is special case of ____ distribution.

A
Theta
B
Alpha
C
Beta
D
Gamma
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 26 Explanation: 
The exponential distribution is not the same as the class of exponential families of distributions, which is a large class of probability distributions that includes the exponential distribution as one of its members, but also includes the normal distribution, binomial distribution, gamma distribution, Poisson, and many others.
Question 27
The algebraic sum of the deviations of all the variables from their mean i.e., is:
A
0
B
1
C
-1
D
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 27 Explanation: 
The sum of the deviations from the mean is zero. This will always be the case as it is a property of the sample mean, i.e., the sum of the deviations below the mean will always equal the sum of the deviations above the mean. However, the goal is to capture the magnitude of these deviations in a summary measure.
Question 28

The mean, mode and median are connected by the empirical relationship:

A
Mean-mode = 2(mean-median)
B
Mean-mode = 3(mean-median)
C
Mean-mode = (mean-mode)/2
D
Mean-mode = (mean-mode)/3
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 28 Explanation: 
A distribution in which the values of mean, median and mode coincide (i.e. mean = median = mode) is known as a symmetrical distribution. Conversely, when values of mean, median and mode are not equal the distribution is known as asymmetrical or skewed distribution. In moderately skewed or asymmetrical distribution a very important relationship exists among these three measures of central tendency. In such distributions the distance between the mean and median is about one-third of the distance between the mean and mode, as will be clear from the diagrams 1 and 2. Karl Pearson expressed this relationship as:
Mode = mean - 3 [mean - median]
Mode = 3 median - 2 mean
and Median = mode + ⅔ [mean-mode]
Question 29

If a random variable takes a finite set of values it is called:

A
Continuous variate
B
Normal variate
C
Discrete variate
D
Exponential variate
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 29 Explanation: 
→ A discrete variable is a variable whose value is obtained by counting.
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:

A
Greatest
B
Positive
C
Least
D
Negative
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 30 Explanation: 
The root mean square deviation when measured from the mean is least value.
Question 31

Portability is not a quality factor of:

A
Software coding
B
Software design
C
Software Process
D
Software testing
       Engineering-Mathematics       Probability       JT(IT) 2016 PART-B Computer Science
Question 31 Explanation: 
McCall’s Quality Factors:
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
How many distinguishable permutations of the letters in the word BANANA are there ?
A
720
B
120
C
60
D
360
       Engineering-Mathematics       Probability       UGC NET CS 2017 Nov- paper-2
Question 32 Explanation: 
Total distinguishable permutations means no repetition.
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
What is the probability that a randomly selected bit string of length 10 is a palindrome?
A
1/64
B
1/32
C
1/8
D
1⁄4
       Engineering-Mathematics       Probability       UGC NET CS 2016 Aug- paper-2
Question 33 Explanation: 
Palindrome is a number that remains the same when its digits are reversed.
Ex: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121...,
Step-1:​ 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​ .
Step-2:​ The probability 2​ 5​ /2​ 10
= 1/2​ 5
= 1/32
Question 34
There are three cards in a box. Both sides of one card are black, both sides of one card are red, and the third card has one black side and one red side. We pick a card at random and observe only one side. What is the probability that the opposite side is the same colour as the one side we observed?
A
3/4
B
2/3
C
1/2
D
1⁄3
       Engineering-Mathematics       Probability       UGC NET CS 2016 July- paper-2
Question 34 Explanation: 
Given data,
-- 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.
Step-1: 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
How many committees of five people can be chosen from 20 men and 12 women such that each committee contains at least three women?
A
75240
B
52492
C
41800
D
9900
       Engineering-Mathematics Engineering-Mathematics       Probability       UGC NET CS 2015 Dec- paper-2
Question 35 Explanation: 
Given data,
-- 20 men and 12 women
-- 5 people can choose from men and women
-- Each committee contains at least three women
Step-1: We have a constraint that each committee contains atleast 3 women.
possibility-1: 2 men + 3 women.
possibility-2: 1 men + 4 women.
possibility-3: 0 men + 5 women.
Step-1: 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
Consider an experiment of tossing two fair dice, one black and one red. What is the probability that the number on the black die divides the number on red die?
A
22 / 36
B
12 / 36
C
14 / 36
D
6 / 36
       Engineering-Mathematics       Probability       UGC NET CS 2015 Jun- paper-2
Question 36 Explanation: 
→ From the given data, there are two dice and each dice the possible ways are 6. So total possible ways are 6*6=36 ways.
→ 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
Let A and B be two arbitrary events, then :
A
P(A∩B) = P(A) P(B)
B
P(P∪B) = P(A) + P(B)
C
P(A∪B) ≤ P(A) + P(B)
D
P(A​ / ​ B) = P(A∩B) + P(B)
       Engineering-Mathematics       Probability       UGC NET CS 2005 Dec-Paper-2
Question 37 Explanation: 
Option-A is happens when A and B are independent.
Option-B is happens when A and B are mutually exclusive.
Option-C is not happens.
Option-D is P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).
Question 38
Consider a set A = {1, 2, 3, ........, 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5 ?
A
533
B
599
C
467
D
66
       Engineering-Mathematics       Probability       UGC NET CS 2014 Dec-Paper-2
Question 38 Explanation: 
Method-1:
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=?
Step-1: To find divisible by 3 numbers are
=⌊1000/3⌋
= 333
Step-2: To find divisible by 5 numbers are
=⌊1000/5⌋
= 200
Step-3: 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+200-66
= 467
Note: We are using floor because excluding fraction value.
Method-2:
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 method-1 ratherhan method-2
Question 39
A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k=1 appears in the printout at least once ?
A
0.5
B
0.704
C
0.632121
D
0.68
       Engineering-Mathematics       Probability       UGC NET CS 2014 Dec-Paper-2
Question 39 Explanation: 
Probability that the value k=1 appears in the printout at least once is
1-p(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
In a multi-user operating system, 30 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 40 minutes, when arrival pattern is a poisson distribution, is .
A
e​ -15
B
1-e​ -15
C
1-e​ -20
D
e​ -20
       Engineering-Mathematics       Probability       UGC NET CS 2018 JUNE Paper-2
Question 40 Explanation: 
→ In probability theory, a Poisson process is a stochastic process that counts the number of events and the time points at which these events occur in a given time interval.
→ The time between each pair of consecutive events has an exponential distribution with parameter λ and each of these inter-arrival times is assumed to be independent of other inter-arrival times.
→ λ = (40*30)/60
=20
P(T>40min)= 1-P(T ≤ 40 min)
=1-(1-e​ -40 min / λ​ )
=1-(1-e​ -40 min / 20​ )
= e​ -20
Question 41
Consider the set of all possible five-card poker hands dealt fairly from a standard deck of fifty-two cards. How many atomic events are there in the joint probability distribution ?
A
2,598,960
B
3,468,960
C
3,958,590
D
2,645,590
       Engineering-Mathematics       Probability       UGC NET CS 2018 JUNE Paper-2
Question 41 Explanation: 
→Joint probability distribution:
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.
Step-1: This problem, we are simply finding combinations of 52 C​ 5
= C(n,r)=C(52,5)
= 52! / (5!(52-5)!)
= 2598960
Question 42
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)(33 − i ) . Let A denote the event that the even number of buffers are full. Then p(A) is:
A
0.515
B
0.785
C
0.758
D
0.485
       Engineering-Mathematics       Probability       UGC NET CS 2018 JUNE Paper-2
Question 42 Explanation: 
Probability of i​ th​ buffer getting full = p(i)=1/562(33−i)
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 (33-0)
P(2) is 1/562 (33-2)
...
P(16) is 1/562 (33-32)
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
A hash table has space for 75 records, then the probability of collision before the table is 6% full.
A
25
B
20
C
35
D
30
E
None of the above
       Engineering-Mathematics       Probability       UGC NET CS 2011 Dec-Paper-2
Question 43 Explanation: 
Given data,
-- Hash table has space =75 slots.
-- For 6% filling it must take =6 slots.
-- Probability of no collision for first 6 entries=?
Step-1: According to given data,
= (75P6) / (756)
= 0.814586387
Step-2: We have to find at least one collision occurs in 6 entries
=1-Probability of no collision for first 6 entries
= 1-0.814586387
= 0.185413613
Note: Given options are wrong. Given option-B is correct answer according to original key.
Question 44
The multiuser operating system, 20 requests are made to use a particular resource per hour, on an average the probability that no request are made in 45 minutes is
A
e–15
B
e–5
C
1 – e–5
D
1 – e–10
       Engineering-Mathematics       Probability       UGC NET CS 2011 Dec-Paper-2
Question 44 Explanation: 
Probability of events for a Poisson distribution:
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 [(λk)/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
What is the probability of choosing correctly an unknown integer between 0 and 9 with 3 chances ?
A
963/1000
B
973/1000
C
983/1000
D
953/1000
E
None of the above
       Engineering-Mathematics       Probability       UGC NET CS 2011 Dec-Paper-2
Question 45 Explanation: 
Probability of getting a number = 1/10
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
In a multi-user operating system, 20 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 45 minutes is
A
e −15
B
e −5
C
1 − e −5
D
1 − e −10
       Engineering-Mathematics       Probability       NIELIT Technical Assistant_2016_march
Question 47
If x and y are independent Gaussian random variables with average value 0 and with same variance, their joint probability density function is given by :
A
p(x, y)=p(x).p(y)
B
p(x, y)=p(x)+p(y)
C
p(x, y)=p(x+y)
D
p(x, y)=p(x).p(y)+p(x)
       Engineering-Mathematics       Probability       UGC NET CS 2009-June-Paper-2
Question 47 Explanation: 
As they are independent, answer could be Option-A, though it need to be checked with zero average too.
Note: Original key given option-C is correct answer
Question 48
A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls are red and two are green ?
A
3/7
B
4/7
C
5/7
D
6/7
       Engineering-Mathematics       Probability       UGC NET CS 2007-Dec-Paper-2
Question 48 Explanation: 


Question 49
Suppose there are n stations in a slotted LAN. Each station attempts to transmit with a probability P in each time slot. The probability that only one station transmits in a given slot is _______.
A
nP(1 – P)n – 1
B
nP
C
P(1 – P)n – 1
D
nP(1 – P)n – 1
       Engineering-Mathematics       Probability       UGC NET CS 2017 Jan- paper-3
Question 49 Explanation: 
This question can be solved by binomial distribution, nC1P1(1-p)n-1 = nP(1-P)n-1
Question 50
Consider a source with symbols A, B, C, D with probability 1/2, 1/4, 1/8, 1/8 respectively. What is the average number of bits per symbol for the Huffman code generated from above information ?
A
2 bits per symbol
B
1.75 bits per symbol
C
1.50 bits per symbol
D
1.25 bits per symbol
       Engineering-Mathematics       Probability       UGC NET CS 2016 July- paper-3
Question 50 Explanation: 




Question 51
Let n = 4 and (a1, a2, a3, a4) = (do, if, int, while). Let p(1 : 4) = (3/8, 3/8, 1/8, 1/8) and Let q(0 : 4) = (2/8, 3/8, 1/8, 1/8, 1/8) where p(i) and q(i) denotes the probability with which we search ai and the identifier x being searched satisfy ai < x < ai+1 respectively. The optimal search tree is given by:
A

B

C

D

       Engineering-Mathematics       Probability       UGC NET CS 2015 Dec - paper-3
Question 51 Explanation: 
The above problem is nothing but Optimal Binary Search Tree(OBST). The OBST we are using the recurrence relation is
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,j-1), 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
Consider the conditional entropy and mutual information for the binary symmetric channel. The input source has alphabet X={0,1} and associated probabilities {1/2, 1/2}. The channel matrix

is where p is the transition probability. Then the conditional entropy is given by:
A
1
B
- plog(p) - (1 - p)log(1 - p)
C
1 + p log(p) + (1 - p)log(1 - p)
D
0
       Engineering-Mathematics       Probability       UGC NET CS 2015 Dec - paper-3
There are 52 questions to complete.
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