September | 2013 | Arbias Hashani

Birthday Problem

September 23, 2013

A pregnant woman is to give birth.

We assume the probability of boy or girl is equal (it is $1/2$ ) and births are independent of each other.

Sometimes assumptions are unrealistic but the ones we have imposed are very realistic. It would be hard to make sense of the question if we didn’t include our assumptions.

Which of the following events is more likely to happen.

Event A: She gives birth to 2 boys and 2 girls.

Event B: She gives birth to 3 boys and 1 girl or 3 girls and 1 boy.

Method 1:

Consider the sample space, let $1$ be a birth for boy and $0$ a birth for girl. We have eight possible combinations for the birth of a boy:

0000
0001
0010
0100
0011
0101
0110
0111

By symmetry we also have eight combinations of birth of a girl, there are total of sixteen combinations. Our sample space consists of the union of these two sets of combinations and has cardinality sixteen (obviously).

For clarification here are the eight combinations for birth of a girl:

1111
1110
1101
1011
1100
1010
1001
1000

The probability $P_1$ of giving birth to two boys and two girls is used by considering the set of the number of combinations that satisfy our event and then taking the cardinality of the set. The combinations

0011
1100
0110
1001
0101
1010

Satisfy our event and by the addition principle we have the probability $P_1$ to be $6/16 = 3/8$ .

The second probability $P_2$ is done in the exactly same way and the combinations

0001
0010
0100
1110
1101
1011
1000
0111

Give us $P_2$ to be $8/16 = 1/2$ .

Therefore event $B$ is more likely.

Method 2:

Let $X$ be the RV counting the number of boys (or girls) born in four Bernoulli trials. $X$ is Binomially distributed with parameters $n=4$ and $p=1/2$. The events $A$ and $B$ can be trivially found.