Urban Workbench: If you are not yet confused, let's look at another family-with-two-children problem! I know that a family has two children. I see one of the children in the mall and notice that she is a girl. What is the probability that both children are girls? Again compare your result with the second part of Example 1.18. Note: Let's agree on what precisely the problem statement means. Here is a more precise statement of the problem: "A family has two children. We choose one of them at random and find out that she is a girl. What is the probability that both children are girls?"

Wednesday, September 1, 2021

If you are not yet confused, let's look at another family-with-two-children problem! I know that a family has two children. I see one of the children in the mall and notice that she is a girl. What is the probability that both children are girls? Again compare your result with the second part of Example 1.18. Note: Let's agree on what precisely the problem statement means. Here is a more precise statement of the problem: "A family has two children. We choose one of them at random and find out that she is a girl. What is the probability that both children are girls?"

Problem

If you are not yet confused, let's look at another family-with-two-children problem! I know that a family has two children. I see one of the children in the mall and notice that she is a girl. What is the probability that both children are girls? Again compare your result with the second part of Example 1.18. Note: Let's agree on what precisely the problem statement means. Here is a more precise statement of the problem: "A family has two children. We choose one of them at random and find out that she is a girl. What is the probability that both children are girls?"

Solution

Here again, we have four possibilities, $G G = (girl, girl), G B, B G, B B$ , and $P (G G) = P (G B) = P (B G) = P (B B) = \frac{1}{4}$ . Now, let $G_{r}$ be the event that a randomly chosen child is a girl. Then we have

P (G_{r} | G G) = 1,

P (G_{r} | G B) = P (G_{r} | B G) = \frac{1}{2},

P (G_{r} | B B) = 0.

We can use Bayes' rule to find $P (G G | G_{r})$ :

$P (G G \| G_{r})$	$= \frac{P (G_{r} \| G G) P (G G)}{P (G_{r})}$
	$= \frac{P (G_{r} \| G G) P (G G)}{P (G_{r} \| G G) P (G G) + P (G_{r} \| G B) P (G B) + P (G_{r} \| B G) P (B G) + P (G_{r} \| B B) P (B B)}$
	$= \frac{1. \frac{1}{4}}{1. \frac{1}{4} + \frac{1}{2} \frac{1}{4} + \frac{1}{2} \frac{1}{4} + 0. \frac{1}{4}}$
	$= \frac{1}{2}$ .

So the answer again is different from the second part of Example 1.18. This is surprising to most people. The two problem statements look very similar but the answers are completely different. This is again similar to the previous problem (please read the explanation there). The conditional sample space here still is $G G, G B, B G$ , but the point here is that these are not equally likely as in Example 1.18.

The probability that a randomly chosen child from a family with two girls is a girl is one, while this probability for a family who has only one girl is $\frac{1}{2}$ . Thus, intuitively, the conditional probability of the outcome $G G$ in this case is higher than $G B$ and $B G$ , and thus this conditional probability must be larger than one third.

Urban Workbench

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Wednesday, September 1, 2021

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