# Conditional probability. under which conditions $E$ and $F$ are independent?

In a village where there are $M$ women and $H$ men, $m$ women smoke and $h$ men smoke. A person is chosen at random. Let $E$ be the event "The chosen person is female" and $F$ the event "the chosen person smokes" under which conditions $E$ and $F$ are independent?

My work:

We know two $E$ and $F$ are independent if

$P(E|F)=\frac{P(E\cap F)}{P(F)}=P(E)$

We need calculate that probability

$P(E\cap F)=\frac{m}{M}$
$P(F)=\frac{mh}{M+H}$
$P(E)=\frac{M}{M+H}$

then

$P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{m(M+H)}{M(mh)}\not =\frac{M}{M+H}$

then the events are dependent.

Here i'm stuck. can someone help me?

• $P(E\cap F)=\frac m{M+H}$ and $P(F)=\frac {m+h}{M+H}$.
– lulu
Apr 21, 2018 at 20:48

The events are independent if and only if $P(E\cap F)=P(E)P(F)$. Thus it has to satisfy the following $$\frac{m}{M+H}=\frac{M}{M+H}\frac{m+h}{M+H} \iff m=\frac{M(m+h)}{M+H}$$
I think the notation is slightly improved if there are $W$ women and $M$ men, $w$ women smoke and $m$ men smoke, $F$ is the event that the chosen person is female, and $S$ is the event that the chosen person smokes. With this notation, events $F$ and $S$ are independent if and only if \begin{align} & P(F \cap S) = P(F) P(S) \\ \iff & \frac{w}{W + M} = \left(\frac{W}{W+M} \right) \left(\frac{w + m}{W + M} \right) \\ \iff & \frac{w}{W} = \frac{w+m}{W + M}. \end{align}