Probability of $\bar{A}$ given B.

Question

Problem: In a high school class there are 10 girls and 8 boys. Of them, 3 girls and 4 boys play chess. If we choose a student at random, determine the probability that he does not play chess knowing that he is a boy.

Hello. The solution is $\frac{1}{2}$, and I can see why but I can't find where is my error in the following:

Let $A = \mbox{play chess}$, $B=\mbox{Its a boy}$, $C=\mbox{Its a girl}$. Then we are being asked $P(\bar{A} / B)$. Using bayes theorem, which states that $P(A / B)=\dfrac{P(B / A) P(A)}{P(B)}$, we have $P(\bar{A} / B) = \dfrac{P(\bar{A})P(B/\bar{A})}{P(B)}$, where $P(\bar{A})=\frac{11}{18}$, $P(B)=\frac{8}{18}$, and $P(B/\bar{A}) = \frac{4}{8}$ which doesn't give $1/2$. Where is my error?

Answer 1

$$p(B/\overline{A}) = \frac{p(\overline{A},B)}{p(\overline{A})} = \frac{\frac{4}{18}}{\frac{11}{18}} = \frac{4}{11}.$$

Plugging in the corrected computation of $p(B/\overline{A})$ gives

$$\frac{\frac{11}{18} \times \frac{4}{11}}{\frac{8}{18}} = \frac{4}{8}.$$

Answer 2

Here is the another method to solve this problem without using girls. We know, that the total amount of boys is 8, 4 boys play chess, therefore, 4 boys don't play chess. $$ \binom{4}{1}\text{ - the quantity of ways to choose a boy, that play chess.} \\ \binom{4}{1}\text{ - the quantity of ways to choose a boy, that don't play chess.} \\ \text{Probability}=\frac{\left(\text{don't play chess}\right)}{\left(\text{don't play chess}\right)+\left(\text{play chess}\right)}=\frac{\frac{\binom{4}{1}}{\binom{18}{1}}}{\frac{\binom{4}{1}}{\binom{18}{1}}+\frac{\binom{4}{1}}{\binom{18}{1}}}=\frac{\frac{4}{18}}{\frac{4}{18}+\frac{4}{18}}=\frac{4}{8}=\frac{1}{2}. $$