This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Set Theory of Probability – 1”.

1. A and B are two events such that P(A) = 0.4 and P(A ∩ B) = 0.2 Then P(A ∩ B) is equal to ___________

a) 0.4

b) 0.2

c) 0.6

d) 0.8

View Answer

Explanation: P(A ∩ B) = P(A – (A ∩ B))

= P(A) – P(A ∩ B)

= 0.6 – 0.2 Using P(A) = 1 – P(A)

= 0.4.

2. A problem in mathematics is given to three students A, B and C. If the probability of A solving the problem is ^{1}⁄_{2} and B not solving it is ^{1}⁄_{4}. The whole probability of the problem being solved is ^{63}⁄_{64} then what is the probability of solving it?

a) ^{1}⁄_{8}

b) ^{1}⁄_{64}

c) ^{7}⁄_{8}

d) ^{1}⁄_{2}

View Answer

Explanation:

Let A be the event of A solving the problem

Let B be the event of B solving the problem

Let C be the event of C solving the problem

Given P(a) =

^{1}⁄

_{2}, P(~B) =

^{1}⁄

_{4}and P(A ∪ B ∪ C) = 63/64

We know P(A ∪ B ∪ C) = 1 – P(A ∪ B ∪ C)

= 1 – P(A ∩ B ∩ C)

= 1 – P(A) P(B) P(C)

Let P(C) = p

ie ^{63}⁄_{64} = 1 – (^{1}⁄_{2})(^{1}⁄_{4})(p)

= 1 – ^{p}⁄_{8}

⇒ P =1/8 = P(C)

⇒P(C) = 1 – P = 1 – ^{1}⁄_{8} = ^{7}⁄_{8}.

3. Let A and B be two events such that P(A) = ^{1}⁄_{5} While P(A or B) = ^{1}⁄_{2}. Let P(B) = P. For what values of P are A and B independent?

a) ^{1}⁄_{10} and ^{3}⁄_{10}

b) ^{3}⁄_{10} and ^{4}⁄_{5}

c) ^{3}⁄_{8} only

d) ^{3}⁄_{10}

view Answer

Explanation: For independent events,

P(A ∩ B) = P(A) P(B)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= P(A) + P(B) – P(A) P(B)

=

^{1}⁄

_{5}+ P (

^{1}⁄

_{5})P

⇒

^{1}⁄

_{2}=

^{1}⁄

_{5}+

^{4}⁄

_{5}P

⇒ P=

^{3}⁄

_{8}.

4. If A and B are two mutually exclusive events with P(~A) = ^{5}⁄_{6} and P(b) = ^{1}⁄_{3} then P(A /~B) is equal to ___________

a) ^{1}⁄_{4}

b) ^{1}⁄_{2}

c) 0, since mutually exclusive

d) ^{5}⁄_{18}

View Answer

Explanation: As A and B are mutually exclusive we have

\(A\cap\bar{B}\)

And Hence

\(P(A/\bar{B})=\frac{P(A\cap\bar{B})}{P(\bar{B})}\)

\(\frac{1-P(\bar{A})}{1-P(\bar{B})}=\frac{1-\frac{5}{6}}{1-\frac{1}{3}}\)

\(P(A/\bar{B})=\frac{1}{4}\)

5. If A and B are two events such that P(a) = 0.2, P(b) = 0.6 and P(A /B) = 0.2 then the value of P(A /~B) is ___________

a) 0.2

b) 0.5

c) 0.8

d) ^{1}⁄_{3}

View Answer

Explanation: For independent events,

P(A /~B) = P(a) = 0.2.

6. If A and B are two mutually exclusive events with P(a) > 0 and P(b) > 0 then it implies they are also independent.

a) True

b) False

View Answer

Explanation: P(A ∩ B) = 0 as (A ∩ B) = ∅

But P(A ∩ B) ≠ 0 , as P(a) > 0 and P(b) > 0

P(A ∩ B) = P(A) P(B), for independent events.

7. Let A and B be two events such that the occurrence of A implies occurrence of B, But not vice-versa, then the correct relation between P(a) and P(b) is?

a) P(A) < P(B)

b) P(B) ≥ P(A)

c) P(A) = P(B)

d) P(A) ≥ P(B)

View Answer

Explanation: Here, according to the given statement A ⊆ B

P(B) = P(A ∪ (A ∩ B)) (∵ A ∩ B = A)

= P(A) + P(A ∩ B)

Therefore, P(B) ≥ P(A)

8. In a sample space S, if P(a) = 0, then A is independent of any other event.

a) True

b) False

View Answer

Explanation: P(a) = 0 (impossible event)

Hence, A is not dependent on any other event.

9. If A ⊂ B and B ⊂ A then,

a) P(A) > P(B)

b) P(A) < P(B)

c) P(A) = P(B)

d) P(A) < P(B)

View Answer

Explanation: A ⊂ B and B ⊂ A => A = B

Hence P(a) = P(b).

10. If A ⊂ B then?

a) P(a) > P(b)

b) P(A) ≥ P(B)

c) P(B) = P(A)

d) P(B) = P(B)

View Answer

Explanation: A ⊂ B => B ⊂ A

Therefore, P(A) ≥ P(B)

11. If A is a perfect subset of B and P(a < Pb), then P(B – A) is equal to ____________

a) P(a) / P(b)

b) P(a)P(b)

c) P(a) + P(b)

d) P(b) – P(a)

View Answer

Explanation: From Basic Theorem of probability,

P(B – A) = P(b) – P(a), this is true only if the condition given in the question is true.

12. What is the probability of an impossible event?

a) 0

b) 1

c) Not defined

d) Insufficient data

View Answer

Explanation: If the probability of an event is 0, then it is called as an impossible event.

13. If A = A_{1} ∪ A_{2}……..∪ A_{n}, where A_{1}…A_{n} are mutually exclusive events then?

a) \(\sum_{i=0}^n P(A_i)\)

b) \(\sum_{i=1}^n P(A_i)\)

c) \(\prod_{i=0}^n P(A_i)\)

d) Not defined

View Answer

Explanation: A = A

_{1}∪ A

_{2}……..∪ A

_{n}, where A

_{1}…A

_{n}

Since A

_{1}…A

_{n}are mutually exclusive

P(a) = P(A

_{1}) + P(A

_{2}) + … + P(A

_{n})

Therefore p(a)=\(\sum_{i=1}^n P(A_i)\)

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