Probability that last child is a boy
Solution 1:
If he has more daughters than sons, Below are the 5 possible cases:
D D D D --> All Daughters
S D D D --> 3 Daughters
D S D D
D D S D
D D D S
So probability of having last child as son is = 1/5.
Solution 2:
The number of girls in the family would have a binomial distribution, so the prior probability that there are 3 or 4 girls in the family would be:
$$\begin{align} \mathsf P(B) & = {4\choose 3}(\tfrac 1 2)^3(\tfrac 1 2)+{4\choose 4}(\tfrac 1 2)^4 \\ & = \frac 5{16} \end{align}$$
Now for the probability that the last child in the family is a boy and that there are more girls than boys in the family is equal to: the prior probability that the first three children are girls and the last is a boy:
$$\begin{align} \mathsf P(A\cap B) & = \frac{1}{16} \end{align}$$
Thus the posterior probability, that the last child is a boy given that their are more girls in the family than boys is:
$$\begin{align} \mathsf P(A\mid B) & = \frac{\mathsf P(A\cap B)}{\mathsf P(B)} \\ & = {\frac 1 {16}}\bigg/\frac 5 {16} \\ & = \dfrac 1 5 \end{align}$$
Solution 3:
There are $2^4=16$ possible permutations of children, e.g. MMMM, or MFFM, or FFFM, or FFFF (here order is important, hopefully it's clear that MFFF means the first child is male, the second is female, and so on) and each is equally likely. Now it just becomes conditional probability.
Let $A$ be the event that the last child is male. Let $B$ be the event that there are more female children than male children. The probability we are looking for is $P(A|B)$, the probability of $A$ given $B$. This is given by $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$ where $A\cap B$ is the event where both $A$ and $B$ occur. Let's look at the terms individually:
- $A\cap B$ occurs precisely when Johnny has more daughters than sons, and the last child is a son. It's not hard to see there's only one possibility here: FFFM. Hence $P(A\cap B)=\frac1{16}$.
- $B$ occurs either when Johnny has no sons (one option, FFFF) or one son (four options, MFFF, FMFF, FFMF and FFFM). There are five options in total, so $P(B)=\frac5{16}$.
Putting this all together we get $$P(A|B)=\frac{\frac1{16}}{\frac5{16}}=\frac15$$ so the probability that Johnny's last child was a son is one in five.