Probability and Statistics Questions and Answers – Set Theory of Probability – 1

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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

Answer: a
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.

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2. A problem in mathematics is given to three students A, B and C. If the probability of A solving the problem is 12 and B not solving it is 14. The whole probability of the problem being solved is 6364 then what is the probability of solving it?
a) 18
b) 164
c) 78
d) 12
View Answer

Answer: c
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) = 12, P(~B) = 14 and P(A ∪ B ∪ C) = 63/64

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

= 1 – P(ABC)

= 1 – P(A) P(B) P(C)

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Let P(C) = p
ie 6364 = 1 – (12)(14)(p)

= 1 – p8
⇒ P =1/8 = P(C)
⇒P(C) = 1 – P = 1 – 18 = 78.

3. Let A and B be two events such that P(A) = 15 While P(A or B) = 12. Let P(B) = P. For what values of P are A and B independent?
a) 110 and 310
b) 310 and 45
c) 38 only
d) 310
view Answer

Answer: c
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)
= 15 + P (15)P
12 = 15 + 45P
⇒ P= 38.

4. If A and B are two mutually exclusive events with P(~A) = 56 and P(b) = 13 then P(A /~B) is equal to ___________
a) 14
b) 12
c) 0, since mutually exclusive
d) 518
View Answer

Answer: a
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) 13
View Answer

Answer: a
Explanation: For independent events,
P(A /~B) = P(a) = 0.2.
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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

Answer: b
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

Answer: b
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

Answer: a
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

Answer: c
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

Answer: b
Explanation: A ⊂ B => BA
Therefore, P(A) ≥ P(B)
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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

Answer: d
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

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

13. If A = A1 ∪ A2……..∪ An, where A1…An 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

Answer: b
Explanation: A = A1 ∪ A2……..∪ An, where A1…An
Since A1…An are mutually exclusive
P(a) = P(A1) + P(A2) + … + P(An)
Therefore p(a)=\(\sum_{i=1}^n P(A_i)\)

Sanfoundry Global Education & Learning Series – Probability and Statistics.

To practice all areas of Probability and Statistics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn