JEE Exam  >  JEE Tests  >  Mathematics (Maths) Class 12  >  Test: Independent Events - JEE MCQ

Test: Independent Events - JEE MCQ


Test Description

10 Questions MCQ Test Mathematics (Maths) Class 12 - Test: Independent Events

Test: Independent Events for JEE 2024 is part of Mathematics (Maths) Class 12 preparation. The Test: Independent Events questions and answers have been prepared according to the JEE exam syllabus.The Test: Independent Events MCQs are made for JEE 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 Mathematics (Maths) Class 12 for JEE & Test: Independent Events solutions in Hindi for Mathematics (Maths) Class 12 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Independent Events | 10 questions in 10 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) Class 12 for JEE 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

204 videos|288 docs|139 tests
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 JEE

204 videos|288 docs|139 tests
Download as PDF

Top Courses for JEE