A city council consists of six people of party $P_1$, two of whom are women, and six of party $P_2$, four of whom are men. A member is chosen at random. If the member chosen is a man, what is the probability that he is from party $P_1$?
Total number of members in city council $= 12$
$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 12$
Total number of men in the council $= 8$; $\;$ total number of women in the council $= 4$
Let $A$ be the event of selecting a member from party $P_1$
Let $B$ be the event that the member chosen is a man
Then, $n \left(B\right) = 8$
$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{8}{12} = \dfrac{2}{3}$
$\left(A \cap B\right) = \;$ event of selecting party $P_1$ AND a man
Party $P_1$ can be selected in $1$ way;
A man from $P_1$ can be selected in $4$ ways
$\therefore \;$ Number of ways of selecting party $P_1$ $\;$ AND $\;$ a man $= 1 \times 4 = 4$ ways
$\therefore \;$ Number of elements in $\left(A \cap B\right) = 4$
$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{4}{12} = \dfrac{1}{3}$
$\therefore \;$ P(selecting party $P_1$ when the member is a man)
$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{1/3}{2/3} = \dfrac{1}{2}$