Conditional Probability question given probability that child has brown hair is 1/4 The probability that a child has brown hair is $\frac{1}{4}$. Assume independence between children. Consider a family with 3 children.
a) If it is known that at least one child has brown hair, what is the probability that at least two children have brown hair?
My solution -
$\frac{P(at least 2 brown hair  \mid  at least 1 brown hair)}{P(at least 1 brown hair)}$ 
= $\frac{(\frac{1}{4})^2}{(\frac{1}{4})^3} = \frac{1}{4}$
Is this correct?
b) If it is known that the youngest child has brown hair, what is the probability that at least two children have brown hair?
I have no idea how to solve this problem when given info about the youngest child.
 A: Your answer for a) is not correct.
The probability that at least one child has brown hair is the complement of the probability that their hair are all not brown:
$$p_1:=1-\left(\frac{3}{4}\right)^3.$$
The probability that at least two children have brown hair is
$$p_2=\underbrace{3\cdot \frac{3}{4}\cdot\frac{1}{4^2}}_{BBN,BNB,NBB}+\underbrace{\frac{1}{4^3}}_{BBB}.$$
So the answer for a) is $\frac{p_2}{p_1}=\frac{10}{37}$ which is greater than $\frac{1}{4}$.
Now for b), we may assume that the first child is the youngest. So
$$p_1=P(\{BNN,BBN,BNB,BBB\})\quad \text{and}\quad
p_2=P(\{BBN,BNB,BBB\}).$$
So what is $\frac{p_2}{p_1}$ in this case?
A: a) your solution is wrong. You should write:
$P(\text{at least two}) = 3C2 \cdot 0.75 \cdot 0.25^2 + 0.25^3$
(multiply because you don't care which child is brown haired)
$P(\text{at least one}) = 1 - 0.75^3$
$P(\text{two when known one}) = \frac{P(\text{at least two})}{P(\text{at least one})$
b) Now, we don't multiply.
$P(\text{youngest brown headed}) = 0.25$
$P(\text{at least two exclude youngest}) = 2 \cdot 0.75 \cdot 0.25^2 + 0.25^3$
$P(\text{answer}) = \dfrac{P(\text{at least two exclude youngest})}{ P(\text{youngest brown headed})}$
