Mrs. Grundy has two children. Given that Mrs. Grundy has at least one child born on a Monday, what is the probability that both her children were born on Mondays?
Assume that each child was equally likely to be born on any day of the week, and that the two birthdays are independent (Mrs. Grundy doesn't have twins!).
I ended up with 4/25 but I was wrong. Thanks!
On
There are $7^2$ ways to write distinct pairs of days, but we're only concerned with those containing at least one Monday.
$7$ of these will have Monday first, and $6$ of the rest will have Monday second. These $13$ are the set of "at least one Monday". Among them is one pair of double Monday.
$$\mathsf P(\text{both on Monday}\mid\text{at least one on Monday}) = \frac{1}{13}$$
Observe that: \begin{align*} \Pr[\text{both Monday} \mid \text{at least one Monday}] &= \frac{\Pr[\text{both Monday and at least one Monday}]}{\Pr[\text{at least one Monday}]} \\ &= \frac{\Pr[\text{both Monday}]}{1 - \Pr[\text{both not Monday}]} \\ &= \frac{(\frac{1}{7})^2}{1 - (\frac{6}{7})^2} \\ &= \frac{1}{49 - 36} \\ &= \frac{1}{13} \end{align*}