Commerce Exam  >  Commerce Tests  >  Test: Independent Events - Commerce MCQ

Test: Independent Events - Commerce MCQ


Test Description

10 Questions MCQ Test - Test: Independent Events

Test: Independent Events for Commerce 2024 is part of Commerce preparation. The Test: Independent Events questions and answers have been prepared according to the Commerce exam syllabus.The Test: Independent Events MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Independent Events below.
Solutions of Test: Independent Events questions in English are available as part of our course for Commerce & Test: Independent Events 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: Independent Events | 10 questions in 10 minutes | Mock test for Commerce preparation | Free important questions MCQ to study for Commerce Exam | Download free PDF with solutions
Test: Independent Events - Question 1

A die is tossed twice. The probability of getting 1, 2, 3 or 4 on the first toss and 4, 5, or 6 on the second toss is:​

Detailed Solution for Test: Independent Events - Question 1

In each case, the sample space is given by S={1,2,3,4,5,6}.
Let E = event of getting a 1, 2, 3 or 4 on the first toss.
And, F = event of getting a 5, 6,or 7 on the second toss.
Then, P(E) = 4/6 = 2/3 
and P(F) = 3/6 = 1/2
Clearly, E and F are independent events.
∴ required probability = P(E∩F) = P(E)×P(F) [∵ E and F are independent]
= 2/3 * 1/2 = 1/3

Test: Independent Events - Question 2

If A and B are two independent events, then P(A ∩ B) =​

Detailed Solution for Test: Independent Events - Question 2

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Independent Events - Question 3

What is the probability of picking a spade from a normal pack of cards and rolling an odd number on a die?​

Detailed Solution for Test: Independent Events - Question 3

Probability of getting odd no. while tossing a die = 3/6 = 1/2
Probability of getting spade from deck of 52 cards = 13/52 = 1/4
Probability of picking a spade from normal pack and rolling odd number = 1/2*1/4 = 1/8

Test: Independent Events - Question 4

If A and B are two independent events, then

Detailed Solution for Test: Independent Events - Question 4

P(A|B) = P(A∩B)/P(B)
=[P(A)P(B)]/P(B) [A is subset of B]
= P(A)

Test: Independent Events - Question 5

Two parts A and B of a machine is manufactured by a firm. Out of 100 A’s 12 are likely to be defective and Out of 100 B’s 8 are likely to be defective. The probability that a machine manufactured by the firm is free from any defect is:​

Detailed Solution for Test: Independent Events - Question 5

 probability of getting good machine part from A = 88/100
probability of getting good machine part from B = 92/100
P(correction) = (88/100)(92/100)
= 506/625

*Multiple options can be correct
Test: Independent Events - Question 6

A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. The probability of occurrence of A is.​

Detailed Solution for Test: Independent Events - Question 6

The probability that both occur simultaneously is 1/6 and the probability that neither occurs is  1/3
Let P(A)=x, P(B)=y
Then P(A)×P(B) = ⅙ becomes xy = 1/6
​And [1−P(A)][1−P(B)]=1/3 becomes (1−x)(1−y)=1/3
​On Solving for x and y, 
we get x=1/3 or x=1/2 which is the probability of occurrence of A.

Test: Independent Events - Question 7

A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. The probability of occurrence of A is.​

Detailed Solution for Test: Independent Events - Question 7

The probability that both occur simultaneously is 1/6 and the probability that neither occurs is  1/3
Let P(A)=x, P(B)=y
Then P(A)×P(B) = ⅙ becomes xy = 1/6
​And [1−P(A)][1−P(B)]=1/3 becomes (1−x)(1−y)=1/3
​On Solving for x and y, 
we get x=1/3 or x=1/2 which is the probability of occurrence of A.

Test: Independent Events - Question 8

Ashmit can solve 80% of the problem given in a book and Amisha can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book?

Detailed Solution for Test: Independent Events - Question 8

Probability that  both Ashmit and Amisha can solve
=0.8×0.7=0.56
Probability that Ashmit  can solve but Amisha not
=0.8×0.3=0.24
Probability that Amisha can solve but Ashmit not 
=0.2×0.7=0.14
so atleast one of them solve the problem
=  0.56+0.24+0.14
=0.94

Test: Independent Events - Question 9

If A and B are independent events, such that P(A ∪ B)= 0.7, P(B) = 0.5, then P(A) = ……​

Detailed Solution for Test: Independent Events - Question 9

 P(A ∪ B) = P(A) + P(B) - P(A)P(B)
= 0.7 = P(A) + 0.5 - 0.5 P(A)
0.2 = 0.5 P(A)
P(A) = ⅖
P(A) = 0.4

Test: Independent Events - Question 10

A student can solve 70% problems of a book and second student solve 50% problem of same book. Find the probability that at least one of them will solve a selected problem from this book.​

Detailed Solution for Test: Independent Events - Question 10

Probability that first and second student can solve
=0.7×0.5=0.35
Probability that first can solve and second cannot solve
=0.7×0.5=0.35
Probability that first cannot solve and Amisha can solve
=0.3×0.5=0.15
Therefore, probability that at least one of them will solve
=0.35+0.35+0.15 = 0.85
=> 85/100
= 17/20

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

Top Courses for Commerce

Download as PDF

Top Courses for Commerce