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This question already has an answer here:

Part A. a company is holding a dinner for working mothers with at least on son. Ms. Smith, a moher of two children is invited. What is the probability that both children are boys?

ANS:P(both boys | at least 1 boy ) = 1/3

Part B. Your new colleague Ms. Parker also have two children. You see her walking with one of her children and that child is a boy. What is the probability that both of them are boys?

ANS: P(both boys | one boy ) = 0.5

I don't understand Part B. If you know one of her children is a boy, does this implies that Ms. Parker has $\geq1$ boy? In mathematical terms, this means that Ms.Parker has at least one boy, true? How come it is different from Part A. Confuse!

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marked as duplicate by JMoravitz, user91500, GNUSupporter 8964民主女神 地下教會, A Blumenthal, Vladhagen Feb 27 '17 at 22:57

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    $\begingroup$ Because in part (A) we don't have any information as to which child was the boy, but in part (B) we have information that the child in our sight specifically is a boy. This is a stronger condition than her having at least one boy since there exist times where her having at least one boy is true but we see a girl instead. $\endgroup$ – JMoravitz Feb 27 '17 at 4:50
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    $\begingroup$ Part A outcome are - boy-boy, boy-girl, girl-boy. So having boy-boy is 1/3. Now, part B outcomes - boy-boy and boy-girl, since first one is definitely a boy. So 1/2. $\endgroup$ – Kaster Feb 27 '17 at 4:58
  • $\begingroup$ Thanks for your reply. I still couldn't see the difference between seeing child A is a boy and just know at least one of them is boy... $\endgroup$ – wrek Feb 27 '17 at 4:58
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    $\begingroup$ Knowing child $A$ is a boy implies that at least one child is a boy. Knowing at least one child is a boy does not imply child $A$ is a boy. Knowing at least one child is a boy is a weaker piece of information $\endgroup$ – JMoravitz Feb 27 '17 at 5:00
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    $\begingroup$ At the risk of confusing you further, part (B) really should have an extra condition in order to ensure the probability is in fact $0.5$ and that is the knowledge that the specific child she is seen walking with was picked uniformly at random from her two children. (the problem changes further if we know something about her selection process for which child to take. For extreme example, if we know that if she has a daughter that she keeps her daughter with her at all times this would imply that since we don't see a daughter with her that she has no daughters) $\endgroup$ – JMoravitz Feb 27 '17 at 5:17
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Let boy be denoted by $B$ and girl by $G$. If a family has two children, then there are the following possibilities for their kids $$BB,BG,GB,GG,$$ each with equal probability, e.g., $P(BB) = P(BG) = 1/4$.

Now, it should be clear for part A, that all we know is that the family is picked randomly from families that are $$BB,BG,GB.$$ Now, the probability that they have two boys is $$P(BB|BB\cup BG \cup GB) = \frac{P(BB\cap (BB\cup BG \cup GB))}{P(BB\cup BG \cup GB)} = \frac{P(BB)}{P(BB\cup BG \cup GB)} = \frac{1/4}{3/4} = \frac{1}{3}.$$

Now, in part B, we have something simlilar, but slightly different. What we have is that we saw a child and it's a boy. Now, the underlying assumption is that we randomly selected a child from the parent's two children. Thus, we have two equally likely scenarios, either we saw child 1 and it was a boy, or we saw child 2 and it was a boy. Let us focus on the former, and note that the probabilities are the same. We note that the event that "child one was a boy" is $BB \cup BG$.

$$P(BB|\text{child one was a boy}) = \frac{P(BB \cap (BB,BG))}{P(BB \cup BG)} = \frac{P(BB)}{P(BB \cup BG)} = \frac{1/4}{2/4} = 1/2.$$ Now, this happened with probability $1/2$ so the total probability that the family in part B has two boys is $$P(BB|\text{(child one was selected AND a boy) OR (child two was selected AND a boy)}).$$ Because it can't happen that child one was selected and child two was selected, these two events are disjoint, hence the probability is equal to $$P(BB|\text{child one was selected AND a boy)} +P(BB|\text{child two was selected AND a boy}) \\ =P(BB|\text{child one was a boy})P(\text{child one was selected}) \\ +P(BB|\text{child two was a boy})P(\text{child two was selected})\\ = \frac{1}{2}\cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}.$$ Where the first equality comes from the fact that whether or not child one was selected was independent of the family composition. Now, this was a bit wordy, but I hope it is helpful in its own right.

Note, throughout it is assumed that you know that a conditional probability of event $A$ given event $B$, denoted $P(A|B)$, is given by $$P(A|B) = \frac{P(A,B)}{P(B)}.$$

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