1. ## conditional probability

Hello, my question is:

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

Thank you for the help in advance!

2. Hello, dagmary!

Suppose that a family has exactly n children (n ≥ 2).
Assume that the probability that any child will be a girl is ½ and that all births are independant.

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

Bayes' Theorem: .$\displaystyle P(B \geq 1\,|\,G \geq 1) \;=\;\frac{P(B \geq 1 \:\cap\:G \geq 1)}{P(G \geq 1)}$

The opposite of "at least boy and at least girl" is "no boys or no girls."
. . $\displaystyle P(\text{no boys}) \:=\:P(\text{all girls}) \:=\:\frac{1}{2^n}$
. . $\displaystyle P(\text{no girls}) \:=\:P(\text{all boys}) \:=\:\frac{1}{2^n}$
Then: .$\displaystyle P(\text{no boys or no girls}) \:=\:\frac{1}{2^n} + \frac{1}{2^n} \:=\:\frac{1}{2^{n-1}}$
. . . Hence: .$\displaystyle P(B \geq 1\:\cap G \geq 1) \:=\:1 - \frac{1}{2^{n-1}} \:=\:\frac{2^{n-1}-1}{2^{n-1}}$

The opposite of "at least one girl" is "no girls".
. . $\displaystyle P(\text{no girls}) \:=\:P(\text{all boys}) \:=\:\left(\frac{1}{2}\right)^n \:=\:\frac{1}{2^n}$
Hence: .$\displaystyle P(G \geq 1) \:=\:1 - \frac{1}{2^n} \:=\:\frac{2^n - 1}{2^n}$

Therefore: .$\displaystyle P(B \geq 1 \:|\:G \geq 1) \;=\;\frac{\dfrac{2^{n-1}-1}{2^{n-1}}} {\dfrac{2^n-1}{2^n}} \;=\;\frac{2(2^{n-1}-1)}{2^n-1} \;=\;\boxed{\frac{2^n-2}{2^n-1}}$