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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 $\frac{20}{63}$.

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    $\begingroup$ 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 $\frac 12$). $\endgroup$
    – lulu
    Mar 12 '17 at 11:38
  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Mar 12 '17 at 11:38
  • $\begingroup$ Did you get this question from todays Score - Allen Career Institute? $\endgroup$ Mar 12 '17 at 11:41
  • $\begingroup$ @lulu I could not understand why it is not (1/2) as there are only two cases . (G,G) and (G,B) $\endgroup$
    – hey
    Mar 12 '17 at 11:44
  • $\begingroup$ @JaideepKhare yes , have you also given that . chat.stackexchange.com/rooms/54160/jee-preparation $\endgroup$
    – hey
    Mar 12 '17 at 11:46
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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}$

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  • $\begingroup$ what mistake I have done $\endgroup$
    – hey
    Mar 12 '17 at 11:52
  • $\begingroup$ how did you find total cases to be 6?They aren't $\endgroup$ Mar 12 '17 at 11:54
  • $\begingroup$ (3G,3b) , (4g,2b) ,(5g,b) ,(6g), (2g,4b) , (1g,5b) $\endgroup$
    – hey
    Mar 12 '17 at 11:57
  • $\begingroup$ Read the Note in my answer ; order matters $\endgroup$ Mar 12 '17 at 11:58
  • $\begingroup$ how do we know about order $\endgroup$
    – hey
    Mar 12 '17 at 11:58
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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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