In a binomial problem where there are n independent trials, 
then the probability of getting exactly x successes and exactly 
n-x failures, where the probability of 1 success in one 
trial is p, is given by the formula:



and the formula for  is



Yours is a binomial problem where there are n = 6 trials, and
we want the probability of getting exactly x = 3 "successes"
(boys) and exactly n-x = 6-3 = 3 "failures" (girls), where the
probability of 1 "success" (boy) in one trial is p = .

The formula for  is



Substituting first in





then in













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Another way to do the problem is:

First calculate the numerator of the probability,
the number of ways to have exactly 3 boys among the
6 children

There are 6 children:

1st-born, 2nd-born, 3rd-born, 4th-born, 5th-born, 6th-born

We want the number of ways we can choose 3 out of these 6 for the
3 boys.  This is "6 choose 3" or 6C3 = 20

Then we calculate the denominator of the probability,
the number of ways to any of the 6 can have either sex.

There are 2 choices for the 1st-born (B or G), then
there are 2 choices for the 2nd-born (B or G), then
there are 2 choices for the 3rd-born (B or G), then
there are 2 choices for the 4th-born (B or G), then
there are 2 choices for the 5th-born (B or G), then
there are 2 choices for the 6th-born (B or G).

That's 

So the desired probability is  or .

Edwin