Q: A couple decides to have $4$ children. If they succeed in having $4$ children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly $2$ girls and $2$ boys?
Explanation of the solution:
Total number of ways of having $4$ kids is $1/2 \cdot 1/2 \cdot 1/2 \cdot 1/2 = 1/16$.
Total number of ways of having exactly $2$ girls and $2$ boys is:
First, count the $2$ girls and $2$ boys as $2$ girls glued together as one and $2$ boys glued together as one so there are $2!$ ways to move them.
Second, count the number of ways the $2$ girls can be moved within themselves $= 2!$
Third, count the number of ways the $2$ boys can be moved within themselves $= 2!$
So total is $2! + 2! + 2! = 6$
So the probability of then is $6/16$ or $3/8$.
What about GIRL, BOY, GIRL, BOY and BOY, GIRL, BOY, GIRL?? I am unable to understand where that was counted?
Best Answer
It wasn't counted, the solution's explanation is erroneous (although the answer is correct). An easier way to compute this is to note that there are 2 choices for each of 4 children, so $2^4=16$ total arrangements. Out of these, $1$ corresponds to all $4$ being boys, $1$ corresponds to $4$ girls, and 4 each of $3$ boys $1$ girl, $3$ girls $1$ boy. Removing these 10 arrangements leaves only arrangements that are $2$ boys $2$ girls, so $16-10=6$. Thus the probability is $6/16$