Commerce Exam  >  Commerce Tests  >  Test: Probability- Case Based Type Questions - Commerce MCQ

Test: Probability- Case Based Type Questions - Commerce MCQ


Test Description

10 Questions MCQ Test - Test: Probability- Case Based Type Questions

Test: Probability- Case Based Type Questions for Commerce 2024 is part of Commerce preparation. The Test: Probability- Case Based Type Questions questions and answers have been prepared according to the Commerce exam syllabus.The Test: Probability- Case Based Type Questions MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Probability- Case Based Type Questions below.
Solutions of Test: Probability- Case Based Type Questions questions in English are available as part of our course for Commerce & Test: Probability- Case Based Type Questions solutions in Hindi for Commerce course. Download more important topics, notes, lectures and mock test series for Commerce Exam by signing up for free. Attempt Test: Probability- Case Based Type Questions | 10 questions in 20 minutes | Mock test for Commerce preparation | Free important questions MCQ to study for Commerce Exam | Download free PDF with solutions
Test: Probability- Case Based Type Questions - Question 1

A child's game has 8 triangles of which 3 are blue and the rest are red, and 10 squares of which 6 are blue and the rest are red. One piece is lost at random.

Q. Find the probability that the lost piece is square.

Detailed Solution for Test: Probability- Case Based Type Questions - Question 1
Number of favourable outcomes for the events that squares is lost, i.e.,

F(E) = 10

T(E) = 8 + 10 = 18

P(getting a square), P(E) = 10/18 = 5/9

Test: Probability- Case Based Type Questions - Question 2

A child's game has 8 triangles of which 3 are blue and the rest are red, and 10 squares of which 6 are blue and the rest are red. One piece is lost at random.

Q. Find the probability that the lost piece is a triangle of red colour.

Detailed Solution for Test: Probability- Case Based Type Questions - Question 2
Number of favourable outcomes for the event that lost figure is triangle of red colour = 5,

i.e., F(E) = 5 and

T(E) = 18

∴ P(lost figure is red triangle), P(E) = 5/18

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Probability- Case Based Type Questions - Question 3

A child's game has 8 triangles of which 3 are blue and the rest are red, and 10 squares of which 6 are blue and the rest are red. One piece is lost at random.

Q. Find the probability that the lost piece is a triangle.

Detailed Solution for Test: Probability- Case Based Type Questions - Question 3
Number of favourable outcomes for the event that lost figure is triangle,

i.e., F (E) = 8

Total figures (square and triangle)

= 8 + 10 = 18

i.e., T(E) = 18

Probability (getting a triangle),

= 8/18 = 4/9

Test: Probability- Case Based Type Questions - Question 4

A child's game has 8 triangles of which 3 are blue and the rest are red, and 10 squares of which 6 are blue and the rest are red. One piece is lost at random.

Q. Find the probability that the lost piece is a square of blue colour.

Detailed Solution for Test: Probability- Case Based Type Questions - Question 4
Number of favourable outcomes for the events that lost figure is square of blue colour,

i.e., F(E) = 6 and T(E) = 18

∴ P(getting a blue square),

=6/18 = 1/3

Test: Probability- Case Based Type Questions - Question 5

A child's game has 8 triangles of which 3 are blue and the rest are red, and 10 squares of which 6 are blue and the rest are red. One piece is lost at random.

Q. How many triangles are of red colour and how many squares are of red colour?

Detailed Solution for Test: Probability- Case Based Type Questions - Question 5

Since, total no. of triangles = 8

Triangles with blue colour = 3

Triangles with red colour = 8 – 3 = 5

and total no. of squares = 10

Squares with blue colour = 6

Squares with red colour = 10 – 6 = 4

Test: Probability- Case Based Type Questions - Question 6

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita

promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John's promotion is the same as that of Gurpreet, Rita's chances of promotion are twice as likely as Johns. Aslam's chances are four times that of John.

Sol. Let Event :

J = John promoted

R = Rita promoted

A = Aslam promoted

G = Gurpreet promoted Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}

i.e., S ={J, R, A, G)

It is given that, chances of John’s promotion is same as that of Gurpreet.

P(J) = P(G)

Rita’s chances of promotion are twice as likely as John.

P(R) = 2P(J)

and Aslam’s chances of promotion are four times that of John.

P(A) = 4P(J)

Q. What is the probability that Aslam got a promotion?

Detailed Solution for Test: Probability- Case Based Type Questions - Question 6
P(Aslam promoted) = 4P(J)

4P(J) = 4 × 1/8 = 1/2

Test: Probability- Case Based Type Questions - Question 7

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita

promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John's promotion is the same as that of Gurpreet, Rita's chances of promotion are twice as likely as Johns. Aslam's chances are four times that of John.

Sol. Let Event :

J = John promoted

R = Rita promoted

A = Aslam promoted

G = Gurpreet promoted Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}

i.e., S ={J, R, A, G)

It is given that, chances of John’s promotion is same as that of Gurpreet.

P(J) = P(G)

Rita’s chances of promotion are twice as likely as John.

P(R) = 2P(J)

and Aslam’s chances of promotion are four times that of John.

P(A) = 4P(J)

Q. If A = {John promoted or Gurpreet promoted}, Find P(A).

Detailed Solution for Test: Probability- Case Based Type Questions - Question 7

A = John promoted or Gurpreet promoted

∴ A = J ∪ G

P(A) = P (J ∪ G)

∴ A = P(J) + P(G) – P (J ∩ G)

A = 1/8 + 1/8 - 0 [∴ P (J ∩ G) = 0]

= 1/4

Test: Probability- Case Based Type Questions - Question 8

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita

promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John's promotion is the same as that of Gurpreet, Rita's chances of promotion are twice as likely as Johns. Aslam's chances are four times that of John.

Sol. Let Event :

J = John promoted

R = Rita promoted

A = Aslam promoted

G = Gurpreet promoted Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}

i.e., S ={J, R, A, G)

It is given that, chances of John’s promotion is same as that of Gurpreet.

P(J) = P(G)

Rita’s chances of promotion are twice as likely as John.

P(R) = 2P(J)

and Aslam’s chances of promotion are four times that of John.

P(A) = 4P(J)

Q. What is the probability that Rita got a promotion?

Detailed Solution for Test: Probability- Case Based Type Questions - Question 8
P(Rita promoted) = P(R)

= 2P(J) = 2 × 1/8 = 1/4

Test: Probability- Case Based Type Questions - Question 9

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita

promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John's promotion is the same as that of Gurpreet, Rita's chances of promotion are twice as likely as Johns. Aslam's chances are four times that of John.

Sol. Let Event :

J = John promoted

R = Rita promoted

A = Aslam promoted

G = Gurpreet promoted Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}

i.e., S ={J, R, A, G)

It is given that, chances of John’s promotion is same as that of Gurpreet.

P(J) = P(G)

Rita’s chances of promotion are twice as likely as John.

P(R) = 2P(J)

and Aslam’s chances of promotion are four times that of John.

P(A) = 4P(J)

Q. What is the probability that Gurpreet got a promotion?

Detailed Solution for Test: Probability- Case Based Type Questions - Question 9
P(Gurpreet promoted) = P(G) = P(J) = 1/8
Test: Probability- Case Based Type Questions - Question 10

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita

promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John's promotion is the same as that of Gurpreet, Rita's chances of promotion are twice as likely as Johns. Aslam's chances are four times that of John.

Sol. Let Event :

J = John promoted

R = Rita promoted

A = Aslam promoted

G = Gurpreet promoted Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}

i.e., S ={J, R, A, G)

It is given that, chances of John’s promotion is same as that of Gurpreet.

P(J) = P(G)

Rita’s chances of promotion are twice as likely as John.

P(R) = 2P(J)

and Aslam’s chances of promotion are four times that of John.

P(A) = 4P(J)

Q. What is the probability that John got a promotion?

Detailed Solution for Test: Probability- Case Based Type Questions - Question 10
Now, P(J) + P(R) + P(A) + P(G) = 1

⇒ P(J) + 2P(J) + 4P(J) + P(J) = 1

⇒ 8P(J) = 1

P(J) = P(John Promoted)= 1/8

Information about Test: Probability- Case Based Type Questions Page
In this test you can find the Exam questions for Test: Probability- Case Based Type Questions solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Probability- Case Based Type Questions, EduRev gives you an ample number of Online tests for practice

Top Courses for Commerce

Download as PDF

Top Courses for Commerce