Question

P(A1) = 0.60, P(A2) = 0.40, P(B | A1) = 0.05, and P(B | A2) = 0.10. Use Bayes’ theorem to determine P(A1 | B).

Answer 1

The result follows from manipulating the conditional probability. By definition,

$P(A_1\mid B)=\dfrac{P(A_1\cap B)}{P(B)}=\dfrac{P(B\mid A_1)P(A_1)}{P(B)}$

$P(A_1\mid B)=\dfrac{0.05\cdot0.60}{P(B)}$

By the law of total probability,

$P(B)=P(B\cap A_1)+P(B\cap A_2)=P(B\mid A_1)P(A_1)+P(B\mid A_2)P(A_2)$

So we have

$P(A_1\mid B)=\dfrac{0.05\cdot0.60}{0.05\cdot0.60+0.10\cdot0.40}\approx0.4286$