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Of a group of children, $0.4$ are boys and $0.6$ are girls. Of the boys, $0.6$ have brown eyes; of the girls, $0.2$ have brown eyes. A child is selected at random from the group.

(a) Find the probability that the child is a girl.

This is $.6$

(b) Find $P(brown eyes | boy)$.

This is $.6$

(c) Find the probability that the child is a boy with brown eyes.

(d) Find the probability that the child is a girl with brown eyes.

(e) Find the probability that the child has brown eyes.

(f) Find the probability that the child is a girl, given that the child has brown eyes.

How can I use $ P(E|F) = P(E∩F) / P(F)$ for the following questions?

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  • $\begingroup$ @LeilaHatami And yet another useless edit to a 3+ years old question. Well done. (Bonus: An inappropriate change of tag.) $\endgroup$ – Did Mar 20 '17 at 16:29
  • $\begingroup$ @Did Your Welcome! $\endgroup$ – MR_BD Mar 20 '17 at 16:29
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Multiply both sides by $P(F)$ to get:

$$ P(E\cap F) =P(E|F)P(F) $$

So for c), for example, you have $$ P(\text{Boy}\cap \text{Brown eyes})=P(\text{Brown eyes}|\text{Boy})P(\text{Boy}) $$

For d), you do it in the same way.

For e), you can use:

$$ P(A)=\sum P(A|B_i)P(B_i) $$

Do you see why?

For f), use the original equation (Bayes' theorem) to find it (just like in c), for example).

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(c) $P\left[\text{boy}\wedge\text{brown eyes}\right]=P\left[\text{brown eyes}\mid\text{boy}\right]P\left[\text{boy}\right]=0.6\times0.4=0.24$

The rule that you are mentioning can also be written as $P\left[E\cap F\right]=P\left[E\mid F\right]P\left[F\right]$.

In fact that is a better way to denote it, because it also stays valid when $P\left[F\right]=0$.

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