# Probability of having three boys and three girls if at least one child is a girl

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 $\frac{20}{63}$.

• The $6$ cases don't have the same probability. Hint; to get started, try to solve the same problem with two children instead of $6$. (the answer is not $\frac 12$).
– lulu
Mar 12 '17 at 11:38
• If you want each case to have equal probability, then the total number of cases is $64-1 = 63$. Or, you could argue that the total number of cases is $6$, but then they do not have the same probability. Mar 12 '17 at 11:38
• Did you get this question from todays Score - Allen Career Institute? Mar 12 '17 at 11:41
• @lulu I could not understand why it is not (1/2) as there are only two cases . (G,G) and (G,B)
– hey
Mar 12 '17 at 11:44
• @JaideepKhare yes , have you also given that . chat.stackexchange.com/rooms/54160/jee-preparation
– hey
Mar 12 '17 at 11:46

$2^6-1$
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}$• what mistake I have done – hey Mar 12 '17 at 11:52 • how did you find total cases to be 6?They aren't Mar 12 '17 at 11:54 • (3G,3b) , (4g,2b) ,(5g,b) ,(6g), (2g,4b) , (1g,5b) – hey Mar 12 '17 at 11:57 • Read the Note in my answer ; order matters Mar 12 '17 at 11:58 • how do we know about order – hey Mar 12 '17 at 11:58 This is a problem in conditional probability: the probability that there are three girls given that there is at least one girl. Let$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}