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)

View Answer

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

View Answer

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)

View Answer

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

View Answer

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

View Answer

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.

6. According to Bayes’s formula P(E)=P(E|F) P(F) + P(E|F^{c})P(F^{c}).

a) True

b) False

View Answer

Explanation: E can be expressed as E = (E ∩ F) U (E ∩ F

^{c})

= 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)

View Answer

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

View Answer

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.

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