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I came across these two examples in Introduction to Probability - Blitzstein and I am confused what exactly is the difference between the two questions.

  1. Mr. Smith has two children. At least one of them is a boy. What is the probability that both the children are boys?
  2. A family has two children. You randomly run into one of the two, and learn that he is a boy. What is the conditional probability that both are boys?

Now, in the first question, we take

  • A: Event that both children turn up to be boys
  • B: Event that at least one of them is a boy.

We need to calculate $P(A|B)$ where $\Omega = \{BB,BG,GB,GG\}$ $$ P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(\{BB\})}{P(\{BB,BG,GB\})} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$ For the second question, we already have seen the first child which is B. The second child can either be B or G and the probability of that happening is $\frac{1}{2}$.

I fail to understand these different answers. In both the questions, the prior information we have is the same i.e. one of the children is B. Why are we getting different answers then?

Siddhartha
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  • In the second case, you know which one of them is a boy for sure. In the first case, any one of them could be boy or both. – Math Lover Dec 18 '21 at 19:19
  • @MathLover But there is no order here. How does it matter? – Siddhartha Dec 18 '21 at 19:21
  • This problem gets everyone. If you have two boys, there are two ways to observe one of them. If you only have one boy, there is only one way to observe him. You will probably do better by googling the boy girl problem. – John Douma Dec 18 '21 at 19:24
  • "order" can be much more than order of birth. Anything that discriminates between the two children is an "ordering" - in your case a family has two children, one that you randomly run into and the other one. – shortmanikos Dec 18 '21 at 19:25
  • Take a different example - say the first question said what it says now but in the second question, you do not randomly run into one rather the question said that the elder one was a boy. What would you say now? – Math Lover Dec 18 '21 at 19:29
  • @MathLover So, if we need to solve the same problem by constructing events, the sample space will remain the same, I suppose. And if A: Both are boys, B: Second one is a boy, then will $A \cap B$ = {BB} and B = {BB,GB}? In this case, we can have $\frac{\frac{1}{4}}{\frac{2}{4}}= \frac{1}{2} $ Is this approach correct? – Siddhartha Dec 18 '21 at 19:30
  • I think the 2nd probability is the confusing one. Look at it this way. Say there are 100 families with 2 kids, 25 of each type (BB, BG, GB, GG). So there are 100 boys from these families. And half of the boys have a brother and half of them have a sister. – TickaJules Dec 18 '21 at 19:31
  • In the second scenario, the probability that the other child (the one you didn't meet) is a boy is $1/2$. In the first scenario, the temptation is to also argue "if one of the children is a boy, then the probability that the other child is a boy is $1/2$". Okay, suppose that they do in fact have two boys. Now answer me this: which of the two boys is the other child then? – Pilcrow Dec 18 '21 at 19:32
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    @Siddhartha yes that is correct approach – Math Lover Dec 18 '21 at 19:38
  • This question has been asked repeatedly on this site. here for instance, though that one is itself a duplicate. – lulu Dec 18 '21 at 19:39
  • @lulu There were three different variations of this question. I understood the first two, got stuck on this one. The one you've referenced is one of the variations that I already understood. – Siddhartha Dec 18 '21 at 19:41
  • @MathLover Thank you very much for clearing this confusion. – Siddhartha Dec 18 '21 at 19:42
  • @lulu Then maybe you should read my question again. In the linked question, it is asked how does order/age matter. In my question, I just couldn't figure out if order was even involved here as I felt both the questions were conditioned on the same event. – Siddhartha Dec 18 '21 at 19:51
  • But that is the source of the "paradox", such as it is. the whole point comes down to "how do you know that the family has at at least one girl?" If you know by specifying one of the kids, that's very different than if you ask "Is at least one of the two kids a girl?" and receiving the answer "Yes." – lulu Dec 18 '21 at 19:54
  • To stress: you can specify the child by, say, actually looking at one (as here) or by referencing the eldest, or the tallest, or by any other distinguishing feature. The point is the same...one of the children is specified and the gender of that particular child is given. – lulu Dec 18 '21 at 19:57
  • @Siddhartha you are welcome! – Math Lover Dec 18 '21 at 20:08

1 Answers1

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Let $T$ be the event that there are two boys, $O$ be the event that we observed a boy and $A$ be the event that there is at least one boy.

Then we have $$P(T|O)=\frac{P(O|T)P(T)}{P(O)}$$

$$P(T|A)=\frac{P(A|T)P(T)}{P(A)}$$

It is easy to see that the numerators in those statements are the same, i.e. $(1)(\frac{1}{4})=\frac{1}{4}$. The difference lies in the denominators.

The probability that we observe a boy is $$P(O)=P(O|BB)P(BB) + P(O|BG)P(BG) + P(O|GB)P(GB) + P(O|GG)P(GG)$$

The probability that there is at least one boy is $$P(A)=P(A|BB)P(BB) + P(A|BG)P(BG) + P(A|GB)P(GB) + P(A|GG)P(GG)$$

Notice the events where there is one boy and one girl. The probability that we observe a boy in those cases is only $\frac{1}{2}$ but the probability that there is at least one boy in those cases is $1$.

Therefore, $P(A)>P(O)$ so $P(T|O)\gt P(T|A)$.

John Douma
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