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Mrs. Grundy has two children. Given that Mrs. Grundy's older child was 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.

The obvious answer would be $\frac{1}{7}$ since the second child is equal likely to be born on any day of the week. But I got this problem out of a high-level math book and that would be too easy. Is $\frac{1}{7}$ the answer? Or am I missing something really big here?

UPDATE: $\frac{1}{7}$ was the answer and I was correct. And no, this is not a duplicate. The other problem has a different part, so whoever marked this question a duplicate was not careful enough to read the whole question.

Yuna Kun
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  • Why can't it be $\frac 17$? – suomynonA Oct 02 '16 at 15:33
  • For your next questions, please turn to more informative titles, possibly taking as an example the modified version of this one. – Did Oct 02 '16 at 15:38
  • @suomynonA I'm not sure if it is the answer, I'm asking for reassurance. – Yuna Kun Oct 02 '16 at 15:38
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    Possible duplicate of Mrs. Grundy's Conditional Probability Problem. That is a pretty exact match, including the name of the Mrs., I wonder how it could have not been proposed while you were asking a question. Quite likely there are older duplicates too, I did not check. – Marc van Leeuwen Oct 02 '16 at 16:05
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    It's not quite an exact match. Here we are asked for the probability given the older child, rather than given at least one child. However, the two questions are very likely to come from the same problem set. If so, then the point of the exercise is to recognise the distinction. – Graham Kemp Oct 02 '16 at 16:16
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    Oops, the problem there is actually different, and it has a different answer. I did find another question that though not identical (it is about sex, not day of the week) has an equivalent set-up as this one. Ironically that question is closed as a duplicate of yet another question of which it is not a duplicate, as it has a different formulation, and consequently a different answer. – Marc van Leeuwen Oct 02 '16 at 16:26

2 Answers2

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The obvious answer is the right one.

To see this formally, let $Y$ be the event that the younger child was born on Monday. Let $O$ be the event that the older child was born on Monday. We are interested in the probability of $Y \cap O$, that both children were born on Monday, given $O$. Using the definition of conditional probability and independence we have

$$P(Y \cap O \mid O) = \frac{P(Y \cap O)}{P(O)} = \frac{P(Y)P(O)}{P(O)} = P(Y) = 1/7$$

aduh
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The answer to this depends how you want to interpret the question. The most basic answer would be $\frac{1}{7}$ and this assumes that the birth of the $2$ children are independent (i.e they don't rely on each other). If you want to be a bit more accurate, you can consider the cases that Mrs Grundy has twins. This would mean (unless we're being really pedantic) that the $2$ children will be born on the same day of the week. This would mean that the days on which her children are born are not independent. According to google roughly $3$% of births result in twins. Let

$A = \{\text{Event Mrs G has twins and they're both born on a Monday}\}$, $B = \{\text{Event Mrs G does not have twins and they're both born on a Monday}\}$ and $C = \{\text{Event Mrs G has at least one child born on a Monday}\}$.

Hence if we take this into account we get the probability $P$ that Mrs G has $2$ children born on a Monday to be \begin{align*} P &= \mathbb{P}((A \cup B)|C)\\ &=\mathbb{P}(A|C)+\mathbb{P}(B|C)\\ &=\frac{\mathbb{P}(A\cap C)}{\mathbb{P}(C)} + \frac{\mathbb{P}(B\cap C)}{\mathbb{P}(C)}\\ &=\frac{\frac{3}{100}\cdot\frac{1}{7}}{1-((\frac{3}{100}\cdot \frac{6}{7})+(\frac{97}{100}\cdot(\frac{6}{7})^2)} + \frac{\frac{97}{100} \cdot (\frac{1}{7})^2}{1-((\frac{3}{100}\cdot \frac{6}{7})+(\frac{97}{100}\cdot(\frac{6}{7})^2)}\\ &=\frac{59}{641} \end{align*} Note I used independence of $A$ and $B$ in the second equality. Hope this helps and you find it an interesting approach to your problem!

Zestylemonzi
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