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This mock test of Test: Independent Events for JEE helps you for every JEE entrance exam.
This contains 10 Multiple Choice Questions for JEE Test: Independent Events (mcq) to study with solutions a complete question bank.
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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:

Solution:

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

QUESTION: 2

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

Solution:

QUESTION: 3

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

Solution:

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

QUESTION: 4

If A and B are two independent events, then

Solution:

P(A|B) = P(A∩B)/P(B)

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

= P(A)

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:

Solution:

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

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.

Solution:

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.

QUESTION: 7

Solution:

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.

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?

Solution:

QUESTION: 9

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

Solution:

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

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.

Solution:

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

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