Question

If in an experiment is conducted, one and only one of the three mutually exclusive events A1, A2, and A3 can occur with these probabilities. P(A1)=0.2 

P (A2)=0.5 and P(A3)=0.3.

The probability of a fourth event B occurring, given that events A1, A2, or A3 occurs are P(B|A1)=0.2     P(B|A2)=0.1      P(B|A3)=0.3

(i). Find P(B). (2.5 points)

(ii). Find P(A1|B). (2.5 points)

(iii). Find P(A2|B). (2.5 points)

(iv). Find P(A3|B). (2.5 points)

Answer 1

Bayes' Theorem: P(A|B)=P(B|A)P(A)/P(B)

A|B means: given event B occurs, then the probability of event A.

Plugging in some numbers we get:

1. P(B|A1)=0.2=P(A1|B)P(B)/P(A1)=P(A1|B)P(B)/0.2; P(A1|B)P(B)=0.04
2. P(B|A2)=0.1=P(A2|B)P(B)/P(A2)=P(A2|B)P(B)/0.5; P(A2|B)P(B)=0.05
3. P(B|A3)=0.3=P(A3|B)P(B)/P(A3)=P(A3|B)P(B)/0.3; P(A3|B)P(B)=0.09
4. (1)/(2): P(A1|B)/P(A2|B)=0.8
5. (1)/(3): P(A1|B)/P(A3|B)=4/9
6. (2)/(3): P(A2|B)/P(A3|B)=5/9

(4)/(5)=P(A3|B)/P(A2|B)=5/9 which is consistent with (6).

We can write P(A2|B)=P(A1|B)/0.8=1.25P(A1|B); P(A3|B)=2.25P(A1|B) and P(B)=0.04/P(A1|B).

We are told that A1, A2 and A3 are mutually exclusive as well as being mutually independent.

P(A1|B)+P(A2|B)+P(A3|B)=1=P(A1|B)(1+1.25+2.25); P(A1|B)=2/9, and P(A2|B)=2/9*1.25=5/18, P(A3|B)=2/9*2.25=1/2=0.5.

P(B)=0.04/(2/9)=9/50=0.18.