Bayes' Theorem problem - Enough information? > There are 5000 British people. A British is either an English, a Scott or a Welsh. 30% of the British are Scottish, and the English are six times the Welsh. The probability that a British has red hair is 0.25, while the probability that a non-Welsh British has red hair is 0.2. What is the expected number of Welsh with red hair? Is it possible to solve this, given the above information? R = the event that a british has red hair. W = the event that a british is welsh. P(R) = 0.5 P(W) = 0.1 P(R|W') = 0.2 I feel like I'm missing information about P(R|W), or am I completely off track? Thanks!

From the given information, $70\%$ are non-Scottish, so the English are $60\%$ and the Welsh are $10\%$.

Let $R$ be the event a person has red hair, and let $W$ be the event a person is Welsh. We want $\Pr(R\mid W)$. We are told that $\Pr(R)=0.25$, and $\Pr(R\mid W')=0.2$.

Solve the problem informally. Out of say $100$ British, we would have $25$ with red hair. Of the $100$ people, $90$ would be non-Welsh, accounting for $18$ red-haired people. So $7$ of the $10$ Welsh would have red hair. Scale up by multiplying by $50$.

**Another way:** One can also use the machinery of conditional probability. Start from $$\Pr(R)=\Pr(R\cap W)+\Pr(R\cap W').\tag{1}$$ To calculate $\Pr(R\cap W')$, use the fact that $$\Pr(R\mid W')=\frac{\Pr(R\cap W')}{\Pr(W')}.\tag{2}$$ You know $\Pr(R|W')$ and $\Pr(W')$, so you can find $\Pr(R\cap W')$. Now fom (1) you can find $\Pr(R\cap W)$, and then, since you know $\Pr(W)$, you can find $\Pr(R\mid W)$.