Question

Three events b1 b2 and b3 are mutually exclusive and exhaustive. If P(b1) = p(b2) = p(b3) and in the context of these events A is define such that p(A/B1) = 0.75 p(A/B2) = 0.93 p(A/B3)
Find p(b1/a ) and p(b3/a)

Answer 1

Use Bayes' rule

Suppose $B_1,B_2,..., B_n$ are mutually exclusive and exhaustive cases (hypotheses) and $A$ is some event (evidence) that has been observed. Then

$$P(B_i|A)={P(B_i)P(A|B_i) \over \displaystyle\sum_{j=1}^nP(B_j)P(A|B_j)}, 1\leq i\leq n$$

$$P(B_1|A)={P(B_1)P(A|B_1) \over P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+P(B_3)P(A|B_3)}$$

$$P(B_2|A)={P(B_2)P(A|B_2) \over P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+P(B_3)P(A|B_3)}$$

$$P(B_3|A)={P(B_3)P(A|B_3) \over P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+P(B_3)P(A|B_3)}$$

Given that

$$P(B_1)=P(B_2)=P(B_3)$$ $$P(A|B_1)=0.75,P(A|B_2)=0.84,P(A|B_3)=0.93$$

We assume that $P(B_1)=P(B_2)=P(B_3)\not=0$. We don't need to compute $P(B_1),P(B_2),P(B_3)$

$$P(B_1|A)={0.75\cdot P(B_1)\over 0.75\cdot P(B_1)+0.84\cdot P(B_2)+0.93\cdot P(B_3)}=$$

$$={0.75\cdot P(B_1)\over 0.75\cdot P(B_1)+0.84\cdot P(B_1)+0.93\cdot P(B_1)}=$$

$$={0.75\over 0.75+0.84+0.93}={25 \over 84}$$

$$P(B_2|A)={0.84\cdot P(B_2)\over 0.75\cdot P(B_1)+0.84\cdot P(B_2)+0.93\cdot P(B_3)}=$$

$$={0.84\cdot P(B_2)\over 0.75\cdot P(B_2)+0.84\cdot P(B_2)+0.93\cdot P(B_2)}=$$

$$={0.84\over 0.75+0.84+0.93}={28 \over 84}$$

$$P(B_3|A)={0.93\cdot P(B_3)\over 0.75\cdot P(B_1)+0.84\cdot P(B_2)+0.93\cdot P(B_3)}=$$

$$={0.93\cdot P(B_3)\over 0.75\cdot P(B_3)+0.84\cdot P(B_3)+0.93\cdot P(B_3)}=$$

$$={0.93\over 0.75+0.84+0.93}={31 \over 84}$$