Probability Test

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Probability Test - 5

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Question 1
1 / -0.25

If A and B are two events of sample space S, then

A
P(A ∩B) = P(B)P(A/B); P(B) ≠0

B
P(A ∪B) = P(A)P(A/B); P(B) ≠0

C
P(A ∪B) = P(B)P(A/B); P(B) ≠0

D
P(A ∩B) = P(A)P(A/B); P(B) ≠0

Solution

The probability of occurrence of event A under the condition that event B has already occurred &P(B)≠0 is called Conditional probability i.e; P(A|B)=P(A ∩ B)/P(B). Multiply with P(B) on both sides implies P(A ∩ B)=P(B).P(A|B). So option 'A' is correct.
Question 2
1 / -0.25

If E, F and G are events with P(G) ≠≠ 0 then P ((E ∪F)|G) given by

A
P (E|G) + P (F|G) –P ((E ∩F)|F)

B
P (E|G) + P (G|F) –P ((E ∩F)|G)

C
P (G|E) + P (F|G) –P ((E ∩F)|E)

D
P (E|G) + P (F|G) –P ((E ∩F)|G)
Question 3
1 / -0.25

If A, B and C are three events of sample space S, then

A
P(A ∩B ∩C) = P(A).P(A/B).P(C/A ∪B)

B
P(A ∩B ∪C) = P(A).P(B/A).P(C/A ∩B)

C
P(A ∪B ∪C) = P(A).P(B/A).P(C/A ∩B)

D
P(A ∩B ∩C) = P(A).P(B/A).P(C/A ∩B)
Question 4
1 / -0.25

If A and B are two events such that P(A) ≠O and P(A) ≠1, then

A

B

C

D

Solution

As we know that, P(A ') = 1 - P(A)
P(A '⋂B ') = P(AUB)'= 1 - P(AUB)
P(A '/B ') = P(A '⋂B ') / P(B ')
⇒( 1 - P(A U B )) / P(B ')
Question 5
1 / -0.25

Let E and F be events of a sample space S of an experiment, then P(E ’/F) = …

A
P(E/F)’

B
P(E ’).P(F)

C
P(E ’)/P(F)

D
1 –P(E/F)
Question 6
1 / -0.25

A fair six-sided die is rolled twice. What is the probability of getting 2 on the first roll and not getting 4 on the second roll?

A
1/36

B
1/18

C
5/36

D
1/6

E
1/3

Solution

The two events mentioned are independent. The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.
P(getting first 2) = 1/6
P(no second 4) = 5/6
Therefore P(getting first 2 and no second 4) = 1/6* 5/6 = 5/36
Question 7
1 / -0.25

In a box containing 100 Ipods, 10 are defective. The probability that out of a sample of 5 Ipods, exactly 1 is defective is:

A

B

C
10^-1

D

Solution

r = 1, n = 5
p = 10/100 = 1/10,
q = 1 - p
q = 1 - 1/10 = 9/10
= nCr (p)^r (q)^{(n - r)}
Exactly one is defective = 5C1 (1/10)¹ (9/10)^{(5 - 1)}
= 5C1 (1/10) (9/10)⁴
= (1/2) (9/10)⁴
Question 8
1 / -0.25

Three coins are tossed. If at least two coins show head, the probability of getting one tail is:

A
3/4

B
1/3

C
1

D
2/3

Solution

Subset={HHH , HHT,HTH,HTT,THH,THT,TTH,TTT}
P(at least two head) = 4/8 = ½
Getting(one tail) = {HHT, HTH, THH}
= 3/8
Therefore, P(one tail) = (⅜) / (½)
= ¾
Question 9
1 / -0.25

If A and B are two events such that P(A) = 0.3 and P(B) = 0.9 and P(B|A) = 0.6,then P(A|B) = ……

A
0.1

B
0.5

C
0.36

D
0.2

Solution

P(B/A) = P(A ∩B)/P(A)
P(A ∩B) = P(B/A))×P(A)
=0.6 ×0.3
=0.18
P(A ∩B) = P(A/B))×P(B)
0.18 = P(A/B) * 0.9
P(A/B) = 0.2
Question 10
1 / -0.25

A bag contains 25 tickets numbered from 1 to 25. Two tickets are drawn one after another without replacement. The probability that both tickets will show even numbers is:

A
11/50

B
11/24

C
13/25

D
3/8

Solution

There are 12 even numbers between 1 to 25
Consider the given events.
A = Even number ticket in the first draw
B = Even number ticket in the second draw
Now, P(A) = 12/25
P(B/A) = 11/24
Required probability : P(A ⋂B) = P(A) * P(B/A)
= (12/25) * (11/24)
= 11/50