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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!

supern3rd
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level – JKnecht Feb 04 '16 at 00:33
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    What exactly did you do to get $4/25$? – Graham Kemp Feb 04 '16 at 02:04

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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*}

Adriano
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How many ways can her children be born? 49.

How many of these births had a Monday? 13.

Of these 13, how many are double Monday? 1.

Answer 1/13.

Dan Barry
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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.

$$\mathsf P(\text{both on Monday}\mid\text{at least one on Monday}) = \frac{1}{13}$$

Graham Kemp
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