Chapter 7 - Section 7.6 - Bayes' Theorem and Applications - Exercises - Page 518: 3

$P(Y|X)=\frac{.8\times.3}{.8\times.3+.5\times .7}\approx .41$

According to Bayes' theorem: $P(Y|X)=\frac{P(X|Y)P(Y)}{P(X|Y)P(Y)+P(X|Y')P(Y')}$ Here, we have $P(X|Y)= .8$ $P(Y)= .3$ $P(X|Y')= .5$ We know, that $P(Y')=1-P(Y)=1-.3=.7$ We can substitute into the definition: $P(Y|X)=\frac{.8\times.3}{.8\times.3+.5\times .7}\approx .41$