Probability Test

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Probability Test - 6

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Question 1
1 / -0.25

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:

A
2/5

B
1/2

C
1/3

D
1/9

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
1 / -0.25

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

A
P(A)/P(B)

B
P(A).P(A | B)

C
P(A).P(B | A)

D
P(A).P(B)

Solution
Question 3
1 / -0.25

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

A
½

B
0

C
¼

D
1/8

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
1 / -0.25

If A and B are two independent events, then

A

B

C

D

Solution

P(A|B) = P(A ∩B)/P(B)
=[P(A)P(B)]/P(B) [A is subset of B]
= P(A)
Question 5
1 / -0.25

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:

A
619/625

B
119/625

C
6/625

D
506/625

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
Question 6
1 / -0.25

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.

A
1/2

B
1/6

C
5/6

D
1/3

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) = ⅙becomes 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
1 / -0.25

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?

A
0.60

B
0.06

C
0.94

D
0.56

Solution

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
Question 8
1 / -0.25

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

A
0.2

B
0.3

C
0.4

D
0.6

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 9
1 / -0.25

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.

A
13/20

B
17/20

C
3/20

D
7/20

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