Answer to Question #166849 in Statistics and Probability for nimesh

Question

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Answer 1

We have that

$P(A)=0.3$

$P(\bar B |A)=0.1$

$P(B|\bar A)=0.2$

Need to find $P(A|B)$ and $P(A|\bar B)$

The probability tree:

$P(\bar A)= 1 - P(A)=1-0.3=0.7$

$P(B|A)=1-P(\bar B|A)=1-0.1=0.9$

$P(\bar B|\bar A)=1-P(B|\bar A)=1-0.2=0.8$

To find $P(A|B)$ and $P(A|\bar B)$ we need to use Bayes theorem:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\bar A)P(\bar A)}=\frac{0.9\cdot 0.3}{0.9\cdot 0.3+0.2 \cdot 0.7}=0.66$

$P(A|\bar B)=\frac{P(\bar B|A)P(A)}{P(\bar B)}=\frac{P(\bar B|A)P(A)}{P(\bar B|A)P(A)+P(\bar B|\bar A)P(\bar A)}=\frac{0.1\cdot 0.3}{0.1\cdot 0.3+0.8 \cdot 0.7}=0.05$