The problem should make clear that P(boy) = P(girl) = 1/2:

A family has 5 children. If each child is equally likely to be a boy or a girl, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

P(exactly n times) = P(one way) * total possible ways.

Let B = boy and G = girl.

P(one way):
One way to get exactly 3 boys and 2 girls is BBBGG.
P(BBBGG) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.

Total possible ways:
Any arrangement of the letters BBBGG will yield exactly 3 boys and 2 girls.
Thus, to account for all the ways to get exactly 3 boys and 2 girls, the result above needs to be multiplied by the number of ways to arrange BBBGG.
Number of ways to arrange BBBGG = 5!/(3!2!) = 10.

Multiplying the results above, we get:
P(exactly 3 boys and 2 girls) = 10 * 1/32 = 10/32 = 5/16.

The correct answer is E.

A family has 5 children. If each child is equally likely to be a boy or a girl, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

Source: www.GMATinsight.com

\[? = P\left( {3B\,\,{\text{and}}\,\,2G\,\,{\text{among}}\,\,5\,\,{\text{children}}} \right)\]

Imagine 3 B´s and 2 G´s in a row... each sequence is associated with the children´s sexuality, from (say) the older to the younger child.

Total number: 2^5 = 32 equiprobable sequences (for instance starting with BBBBB and ending with GGGGG.)

Favorable number: C(5, 3) = 10 , because we must choose among the 5 positions in the sequence, 3 of them to "put" the B´s (and the remaining 2 to "put" the G´s).

\[? = \frac{{C\left( {5,3} \right)}}{{32}} = \frac{{10}}{{32}} = \frac{5}{{16}}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.