Solution: 2014 Winter Final - 21

Author: Michiel Smid

Question

When a couple has a child, this child is a boy with probability 1/2 and a girl with probability 1/2, independent of the gender of previous children. A couple stops having children as soon as they have 2 girls or 2 boys. Define the random variables

G = the number of girls the couple has

and

B = the number of boys the couple has.

Are $G$ and $B$ independent random variables?

(a)

(b)

Yes

Solution

Oh, we just need to find out something like $ Pr(G = 1, B = 1) = Pr(G = 1) \cdot Pr(B = 1) $

Let S be the set of all subsets
$ S = { (BB), (GG), (BGB), (BGG), (GBG), (GBB) } $
$ |S| = 6 $
Let's determine $ G = 1 $
$ (G=1) = { (BGB), (GBB) } $
$ |G=1| = 2 $
$ Pr(G=1) = \frac{2}{6} = \frac{1}{3} $
Let's determine $ B = 1 $
$ (B=1) = { (BGB), (GBG) } $
$ |B=1| = 2 $
$ Pr(B=1) = \frac{2}{6} = \frac{1}{3} $
Let's determine $ G = 1, B = 1 $
There is never a case where there is only 1 boy and 1 girl. We need them to pop out another baby to break the gender tie
$ G=1 \cap B=1 = emptyset $
$ Pr(G=1, B=1) = 0 $

Now, let’s check for independence

$ Pr(G=1 \cap B=1) = Pr(G=1) \cdot Pr(B=1) $

$ 0 = \frac{1}{3} \cdot \frac{1}{3} $

$ 0 = \frac{1}{9} $

$ 0 \neq \frac{1}{9} $

Therefore, $ G $ and $ B $ are not independent random variables.