Solution: 2014 Fall Final - 19

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 a child that has the same gender as their first child. Define the events

A = "the second child is a boy"

and

B = "the couple has at least three children and the third child is a boy".

Which of the following is true?

(a)

None of the above.

(b)

The events $A$ and $B$ are independent.

(c)

The events $A$ and $B$ are not independent.

Solution

We need to determine if events $ A $ and $ B $ are independent.

$ Pr(A) = \frac{1}{2} $
If they have at least 3 kids and the third child is a boy, let's calculate what's needed to get there
Boy, Girl, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $
Girl, Boy, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $
$ Pr(B) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} $
$ Pr(A \cap B) $ now
If the second child is a boy and we want them to keep pumping out babies, then the first two children must be of different genders
That leaves us with Girl, Boy, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $
$ Pr(A \cap B) = \frac{1}{8} $

Now, let’s check if $ Pr(A) \cdot Pr(B) = Pr(A \cap B) $

$ \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} $

$ \frac{1}{8} = \frac{1}{8} $

The events $ A $ and $ B $ are independent. Thus, the statement is true.