Question 1157709
A family has 6 children. Assume that each child is as likely to be a boy as it is to be a girl. Find the 
probability that the family has 6 girls if it is known the family has at least one girl.
<pre>
                               P(all girls AND at least 1 girl)     P(all girls)
P(all girls|at least 1 girl) = -------------------------------- = ------------------
                                     P(at least 1 girl)           P(at least 1 girl)


The oldest could be a boy or girl, that's 2 ways.
The next to oldest could be a boy or girl, that's 2 ways.
The 3rd to oldest could be a boy or girl, that's 2 ways. 
The 3rd to youngest could be a boy or girl, that's 2 ways.
The next to youngest could be a boy or girl, that's 2 ways.
The youngest could be a boy or girl, that's 2 ways.

That's 2∙2∙2∙2∙2∙2 = 2<sup>6</sup> = 64 ways the 6 children could be.

P(all girls) = 1/64
P(at least 1 girl) = 1 - P(all boys) = 1 - 1/64 = 63/64

                                  P(all girls)       1/64
P(all girls|at least 1 girl) = ------------------ = ------- = (1/64)(63/64) = 1/63
                               P(at least 1 girl)    63/64

Edwin</pre>