Given that the family has at least one girl, determine the probability that the family has at least one boy.

Suppose that a family has exactly n children (n ≥ 2). Assume that the probability that any child will be a girl is 1/2 and that all births are independent. Given that the family has at least one girl, determine the probability that the family has at least one boy.

This is how I went about it:

Since we know that the family has at least one girl, we now have to determine the probability that the family has at least one boy from a pool of n-1 children.

Thus, probability is equal to:

1-Probability of having no boys

=1-(1/2)^(n-1)

But the answer is (1-(1/2)^(n-1)) / (1-(1/2)^(n))

The structure of the answer suggests that Bayes theorem is being used in the correct answer, but even the Bayes theorem should lead to my answer. I do not know where I am going wrong?

• Notice the difference between the family having at least one girl, and the family's oldest child being a girl.
– N.U.
May 18 '13 at 13:02

Define $Y$ to be the numbe of boys, $X$ the number of girls. You need: $$P(Y \geq 1 |X \geq 1)=1-P(Y=0|X \geq 1) = 1- \frac{P(Y=0|X \geq 1)P(X \geq 1)}{P(X \geq 1)}\\ =1 -\frac{P(Y=0 \cap X \geq 1)}{1-P(X=0)}=1-\frac{\frac{1}{2^n}}{1-\frac{1}{2^n}}=\frac{1-\frac{1}{2^{n-1}}}{1-\frac{1}{2^n}}$$