Mr A has six children and at least one child is a girl , what is the probability that Mr A has $3$ boys and $3$ girls?
my try
Total cases are $6$
so probability should be $\frac{1}{6}$
but the answer is $\frac{20}{63}$.
$\endgroup$ 92 Answers
$\begingroup$The order of ages of children matters.
Total cases : Since each child can be either girl or boy ; total cases will be
$2^6-1$
I have subtracted one for that case where each baby is a boy.
Favourable cases : We want $3$ child to be Boys and $3 $ girls.Since order does matter, we will have to permute $3 $ Boys and $3₹ Girls, i.e.
$$\frac {6!}{3!3!}=20$$
Probability = Favourable cases/Total cases
Probability $ = \dfrac {20}{63}$
$\endgroup$ 9 $\begingroup$This is a problem in conditional probability: the probability that there are three girls given that there is at least one girl. Let $G$ be the total number of girls in the family. Then, by the definition of conditional probability, $$\begin{align} \Pr(G=3 \;|\; G \ge 1) &= \frac{\Pr((G=3) \; \cap \; (G \ge 1))}{\Pr(G \ge 1)} \\ &= \frac{\Pr(G=3)}{\Pr(G \ge 1)} \\ &= \frac{\Pr(G=3)}{1-\Pr(G = 0)} \\ &= \frac{\binom{6}{3} (1/2)^6}{1-(1/2)^6} \end{align}$$
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