Stat400 HW4 Due Feb 16


Section 3.2 -13,26 Parts (e) and (f) of 13 are tricky, if 2 lines are in use how many are not in use.
Section 3.3 - 32,44
Section 3.4- 49,52

Extra Problem

2. The Boy or Girl Paradox

Suppose a friend of yours is a married woman with two children, not twins. The answers to (i), (ii), (ii) and (iv) are all different.
(i) If you are given that she has at least one girl what is the probability that she has two girls (draw the conditional sample space).

(ii) If you are given that her first child is a girl what is the probability that she has two girls (draw the conditional sample space - it is different).

Now things get really tricky. You don't need Bayes' Theorem for (i) and (ii) - just draw the conditional sample spaces, but you definitely need to use Bayes' Theorem for (iii) and (iv) .

(iii)Suppose you ask her "do you have a girl child" and she says "yes". What is the probability both of her children are girls?
Now you have to compute P(GG| she tells you one of her children is a girl).

Here is how to do the problem.
ou have to use Bayes' Theorem. You will need to compute the right-hand side of Bayes' Theorem, namely:
[P(She tells you one of her children is a girl|GG) P(GG) ] / [ P(She tells you one of her children is a girl|BB) P(BB) + P(She tells you one of her children is a girl|BG) P(BG) + P(She tells you one of her children is a girl|GB) P(GB)+ P(She tells you one of her children is a girl|GG) P(GG)]

(iv) What if she tells you instead "my first child is a girl". What is the probability now that she has two girls?
You have to use Bayes' Theorem again. See Wikipedia "Boy or Girl Paradox" (though this Wikipedia article is not very clearly written).