Dr. Jitesh J. Thakkar

Assignment 2

(Solution) Probability and Statistics

Dr. Jitesh J. Thakkar

Department of Industrial and Systems Engineering

Indian Institute of Technology Kharagpur

Instruction

• Total No. of Questions: 15. Each question carries one point.

• All questions are objective type. In some of the questions, more than one answers are correct.

• This assignment includes true/false statement questions.

Dr. Jitesh J. Thakkar, IIT Kharagpur

Question 1

Bag I contains 6 white and 4 black balls while another

Bag II contains 3 white and 4 black balls. One ball is

drawn at random from one of the bags and it is found to

be black. What is the probability that it was drawn from

Bag II?

a) 0.404

b) 0.583

c) 0.588

d) 0.644


Solution 1

Answer: c (0.588)

Let E1 be the event of choosing the bag I, E2 the event of choosing

the bag II and A be the event of drawing a black ball.

Then, P(E1) = P(E2) = 1

2

Also, P(A|E1) = P(drawing a black ball from Bag I) = 4

10 =

2

5

P(A|E2) = P(drawing a black ball from Bag II) = 4

7

By using Bayes’ theorem, the probability of drawing a black ball

from bag II out of two bags,

P(E2|A) = P(E2)P(A|E2)

P(E1)P(A│E1)+P(E2)P(A|E2) =

1

2×4

71

2×2

5+1

2×4

7

=10

17= 0.588


Question 2

a) 1-c, 2-b, 3-a

b) 1-b, 2-a, 3-c

c) 1-b, 2-c, 3-a

d) 1-c, 2-a, 3-b

Match the following.


1. Events a. Head will be mapped to 1, and tail will

be mapped to 0, in Coin tossing.

2. Sample Space b. The possible outcomes of a stochastic or

random process

3. Random Variable c. The set that consists of all the outcomes

Solution 2

Answer: c (1-b, 2-c, 3-a)


1. Events b. The possible outcomes of a stochastic or

random process

2. Sample Space c. The set that consists of all the outcomes

3. Random Variable a. Head will be mapped to 1, and tail will be

mapped to 0, in Coin tossing.

Question 3

a) 0.22

b) 0.34

c) 0.45

d) None of these

At Cornell School, all first year students must take

chemistry and math. Suppose 25% fail in chemistry, 18%

fail in math, and 9% fail in both. Suppose a first year

student is selected at random. What is the probability that

student selected failed at least one of the courses?


Solution 3

Answer: b (0.34)

To Solve this problem use Additive law of Probability.

𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)

P(C) = 0.25

P(M) = 0.18

𝑃 𝐶 ∩𝑀 = 0.09

P(at least one) = P(C or M) = P(𝐶 ∪𝑀) =

0.25+0.18-0.09 = 0.34


Question 4

a) The probability of rain today is 23%.

b) The probability of rain today is -10%.

c) The probability of rain today is 120%.

d) The probability of rain or no rain today is 80%.

Which of the following statement is correct?


Solution 4

Answer: a (The probability of rain today is 23%)

Remember !!!!

𝑷 𝒂𝒏 𝒆𝒗𝒆𝒏𝒕 ≥ 𝟎

𝑷 𝒂𝒏 𝒆𝒗𝒆𝒏𝒕 ≤ 𝟏

𝑷 𝑺𝒂𝒎𝒑𝒍𝒆 𝑺𝒆𝒕 = 𝟏


The given table shows the data of Two-way Tables and

Probability of a class. Find the probability that the student getting

Ex grade is a girl.

a) 13

36

b) 4

9

c) 5

36

d) 5

13

Question 5


Got Ex Got < Ex

Girl 25 40

Boy 55 60

Answer: d (5

13)

• Total Boy = 115

• Total Girl = 65

• Probability that the student is a girl =P(G)= 65

180

• Total Ex grade holders = 80

• Probability that a student is a Ex grade holder = P(A) = 80/180

• The number of girls getting Ex grade = 25

• Probability that a student getting Ex grade and a girl =

P(A∩G) = 25

180

• Probability that a student getting Ex grade is a girl = P(A|G)

= 𝑃(𝐴∩𝐺)

𝑃(𝐺)=25/180

65/180=25

65=

5

13

Solution 5


A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, NOT replaced and then another pen is drawn. What is the probability of drawing 2 blue pens and 1 black pen?

Question 6

a)1

14

b)2

14

c)3

14

d)4

14


Answer: a (1

14)

Probability of drawing 1 blue pen = 4/9

Probability of drawing another blue pen = 3/8

Probability of drawing 1 black pen = 3/7

Probability of drawing 2 blue pens and 1 black

pen = 4/9 * 3/8 * 3/7 = 1/14

Solution 6


Question 7

If two events are Mutually Exclusive then which of the following holds good?

a) Events whose occurrence do not depend on the

occurrence of any other events

b) The occurrence of one precludes the occurrence of

the other

c) 𝑃 𝐴 ∩ 𝐵 = 0

d) 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 + 𝑃(𝐵)


Solution 7

Answer: b and c [𝑃 𝐴 ∩ 𝐵 = 0] Examples of Mutually Exclusive Events:

1. In a soccer game scoring no goal (event A) and scoring exactly one goal (event B).

2. In a deck of cards Kings (event A) and Queens (event B).

Examples Not Mutually Exclusive Events:

In a deck of cards Kings (event A) and Hearts (event B). They have common cards i.e. Heart King. So 𝐴 ∩ 𝐵 ≠ 0.

Events whose occurrence do not depend on the occurrence of any other events are called independent events.


Question 8 If a coin is tossed five times then what is the

probability that you observe at least one head?

a) 1

32

b) 15

32

c) 23

32

d) 31

32


Solution 8

Answer: d (31

32)


Consider solving this using complement.

Probability of getting no head = P(all tails) = 1/32

P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

Question 9 A desk lamp produced by The Luminar Company was found to be defective (D). There are three factories (A, B, C) where such desk lamps are manufactured. A Quality Control Manager (QCM) is responsible for investigating the source of found defects. This is what the QCM knows about the company's desk lamp production and the possible source of defects:


Factory % of total

production

Probability of defective

lamps

A 0.35 = P(A) 0.015 = P(D | A)

B 0.35 = P(B) 0.010 = P(D | B)

C 0.30 = P(C) 0.020 = P(D | C)

Question 9


a) 0.407

b) 0.381

c) 0.326

d) 0.237

The QCM would like to answer the following

question: If a randomly selected lamp is defective,

what is the probability that the lamp was

manufactured in factory B?

Solution 9

Answer: d (0.237)

Using conditional probability and Bay’s Theorem

P(B|D)=𝑃(𝐵∩𝐷)

𝑃(𝐷)=𝑃 𝐷 𝐵 .𝑃(𝐵)

𝑃(𝐷)

=𝑃 𝐷 𝐵 . 𝑃(𝐵)

𝑃[(𝐷 ∩ 𝐴) ∪ (𝐷 ∩ 𝐵) ∪ (𝐷 ∩ 𝐶)]

=𝑃 𝐷 𝐵 . 𝑃(𝐵)

𝑃 𝐷 𝐴 . 𝑃 𝐴 + 𝑃 𝐷 𝐵 . 𝑃 𝐵 + 𝑃 𝐷 𝐶 . 𝑃(𝐶)

=(0.01). (0.35)

(0.015). (0.35) + (0.01). (0.35) + (0.02). (0.30)= 0.237


Question 10

One card is randomly drawn from a pack of 52

cards. What is the probability that the card drawn

is a face card(Jack, Queen or King)?

a) 1

13

b) 2

13

c) 3

13

d) 4

13


Solution 10

Answer: c (3

13)

Total number of cards, n(S) = 52

Total number of face cards, n(E) = 12

P(E) = n(E)n(S)

=1252=313


Question 11

Consider the experiment of tossing a coin twice.

A is an event having sample set as {HT,HH} and

B is an event having sample set {HT}. Event A is

independent from Event B.

a) True

b) False


Solution 11

Answer: b (False) When two events are said to be independent of each

other, what this means is that the probability that

one event occurs in no way affects the probability of the

other event occurring.


Question 12

In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

a) 0.34

b) 0.53

c) 0.67

d) 0.74


Solution 12

Answer: c (0.67)

P(M and S) = 0.40

P(M) = 0.60

P(S|M) = P(M and S)/P(M) = 0.40/0.60 = 2/3 = 0.67


Question 13

At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remainder are Lorries. If every vehicle is equally likely to leave, find the probability of car leaving second if either a lorry or van had left first.

a)3

10

b)20

33

c)1

10

d)3

5


Solution 13

Answer: b (20

33)

Let S be the sample space and A be the event of a van leaving first.

n(S) = 100 , n(A) = 30

Probability of a van leaving first: P(A) =30

100=

3

10

Let B be the event of a lorry leaving first.

n(B) = 100 – 60 – 30 = 10

Probability of a lorry leaving first: P(B) =10

100=

1

10

If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars. Let T be the sample space and C be the event of a car leaving second.

n(T) = 99, n(C) = 60

Probability of a car leaving after a lorry or van has left: P(B) =60

99=20

33


Question 14 What is the formula to find skewness of a distribution?

a) 𝑔 = (𝑥−𝑥𝑖)

3𝑛𝑖−1

(𝑛−1)𝑠3 Where x is the

observation, 𝑥𝑖 is the mean, n is the total

number of observations and s is the

variance.

b) Skewness = 3×(𝑚𝑒𝑎𝑛−𝑚𝑒𝑑𝑖𝑎𝑛)

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛

c) Both a and b

d) None of these Dr. Jitesh J. Thakkar, IIT Kharagpur

Solution 14 Answer: c (Both a and b)

• 𝑆𝑘𝑒𝑤𝑛𝑒𝑠𝑠(𝑔) = (𝑥−𝑥𝑖)

3𝑛𝑖−1

(𝑛−1)𝑠3 Where x is the

observation, 𝑥𝑖 is the mean, n is the total number of observations and s is the variance.

• The formula to find skewness manually

Skewness = 3×(𝑚𝑒𝑎𝑛−𝑚𝑒𝑑𝑖𝑎𝑛)

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛

We have the following rules as per Bulmer in the year 1979:

• If the skewness comes to less than -1 or greater than +1, the data distribution is highly skewed

• If the skewness comes to between -1 and −1

2 or between +

1

2 and +1, the data distribution is moderately

skewed.

• If the skewness is between − 1

2 and +

1

2, the distribution is approximately symmetric


Question 15

What is the median and mode for the following

list of values?

13, 18, 13, 14, 13, 16, 14, 21, 13

a) 14 and 13

b) 15 and 16

c) 14 and 15

d) 13 and 16


Solution 15

Answer: a (14 and 13)

• The median is the middle value, so to rewrite the list in order:

13, 13, 13, 13, 14, 14, 16, 18, 21

• There are nine numbers in the list, so the middle

one will be 9+1

2 =5

= 5th number, So the median is 14.

• The mode is the number that is repeated more often than any other, so 13 is the mode.


Thank you

Dr. J

itesh

J. T

ha

kka

r, IIT K

ha

ragp

ur

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