Probability Questions and Answers – Marginalization, Independence, Conditioning and Bayes Rule

This set of Probability Multiple Choice Questions & Answers (MCQs) focuses on “Marginalization, Independence, Conditioning and Bayes Rule”.

1. Two events A and B are said to be independent if ______
a) P(EF)=P(E+F)
b) P(EF)=P(E)
c) P(EF)=P(E)P(F)
d) P(EF)=P(F)

Explanation: For two events to be independent the one and only condition is that either P(E+F)=P(E)+ P(F) or P(EF)=P(E)P(F).

2. Two coins are flipped and all four outcomes are assumed to be equally likely. The events E and F are :
E: the first coin gives head
F: the second coin gives tail
Here E and F are ______
a) dependent
b) mutually exclusive
c) equally likely
d) independent

Explanation: P(E)=P({(H,H),(H,T)})=1/2,
(F)=P({(H,T),(T,T)})=1/2,
P(EF)=P({(H,T)})=1/4.

3. Three events E, F, G are said to be independent if _______
a) P(EFG) = P(E)P(F)P(G)
b) P(EF) = P(E)P(F)
c) P(EG) = P(E)P(G)
d) P(EFG) = P(E)P(F)P(G), P(EF) = P(E)P(F), P(EG) = P(E)P(G), P(FG) = P(F)P(G)

Explanation: All the events should be individually independent to other events e.g. E should be independent to G, E should be independent to F, F should be independent to G.

4. Paul is undecided as to whether to take a French course or a Chemistry course. He estimates that his probability of receiving an A grade would be ½ in a French course and 2/3 in a chemistry course. If Paul decides to take his decisions based on the flip of a fair coin, what is the probability he gets an A in chemistry?
a) 2/3
b) 6/8
c) 3/5
d) 1/3

Explanation: C: Paul takes chemistry
A: Paul gets an A
P(CA)
=P(C)P(A|C)
=1/2 x 2/3
=1/3.

5. Suppose that a new policyholder has an accident within a year of enrolling in a policy. What is the probability that he or she is accident prone?
a) 6/15
b) 9/13
c) 6/13
d) 1

Explanation: P (A|B) = P (A∩B)/P (B)
=$$\frac{P(A)P(B|A)}{(P(B)}$$
= (0.3*0.4)/0.26 = 6/13.
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6. According to Bayes’s formula P(E)=P(E|F) P(F) + P(E|Fc)P(Fc).
a) True
b) False

Explanation: E can be expressed as E = (E ∩ F) U (E ∩ Fc)
= P (E|F) P (F) + P (E|F) P (F).

7. Let the events E and F denote the following events
E: the sum of numbers on two dies is 8
F: the number on the first die is 3
When two fair dice are rolled, then the conditional probability of E given that F has already occurred is given by ______
a) P(E)
b) P(F|E)
c) P(F)
d) P(E|F)

Explanation: The event F has already occurred. Now, we have to find the probability of event E happening, with prior knowledge that F has already occurred. P (sum of numbers on two dies is 8 given that the number on the first die is 3) = P (E|F).

8. For two events E and F, P(E|F)=\frac{P(EF)}{(P(F)} holds only if ___________
a) P(F) ≠ 0
b) P(F) ≥ 0
c) P(F) > 0
d) P(F) ≫ 0

Explanation: For a probability to be valid enough the denominator should be strictly greater than 0, cause 0<P(F)≤1, else the probability will be undefined, since any physical quantity divided by zero is undefined.

Sanfoundry Global Education & Learning Series – Probability and Statistics.