Given $Pr(C|M)$ and $Pr(C|F)$, find $Pr(M|C)$ My question: The probability that a male is colourblind is 0.05 while the probability that a female is colourblindis 0.0025. If there is an equal number of males and females in a population, find the probability that a personselected at random from the population is male given that the person is colourblind.
Textbook's answer: $0.9524$
I have managed to solve this with a two-way table.
$Pr(M|C) = \frac {0.05}{0.0525}$
However I am unable to solve the problem using $Pr(M|C) = Pr(M ∩ C)/Pr(C)$
$Pr(C) = Pr(C|M) + Pr(C|F)$
$Pr(C) = 0.0525$
By I am confused of why $Pr(M ∩ C) = Pr(M)$
Intuitively, I know that we are selecting a man from the sample; however formula-vise, it is bit counterintuitive how the probability of selecting a man and a colourblind person is equal to the probability of selecting a man.
It would be appreciated, if you could correct my understanding.
 A: It appears your confusion arises from couple arithmetic errors that come before it.
I assumed that you have obtained that $Pr(M|C) = \frac{0.05}{0.0525}$, and calculated that $Pr(C) = 0.0525$. Thus, by the correct definition of conditional probability $Pr(M|C) = \frac{Pr(M \cap C)}{Pr(C)}$, you are trying to reason that $Pr(M \cap C) = 0.05$, and wonder why it is equal to $Pr(M)$.
The more obvious error that occurred is that $Pr(M)$ is not $0.05$, but actually $0.5$.
The more subtle error that lead to the confusion is that $Pr(C)$ is actually not $0.0525$. You have effectively double counted both the male and female who are colour blind in your population by writing $Pr(C) = Pr(C|M) + Pr(C|F)$. By the law of total probability, it should be
$$Pr(C) = Pr(C|M)Pr(M) + Pr(C|F)Pr(F) = 0.05 \cdot 0.5 + 0.0025 \cdot 0.5 = 0.02625.$$
You can then calculate $Pr(M \cap C)$ using the given $Pr(C|M)$ and $Pr(M)$, and verify the fraction $\frac{Pr(M\cap C)}{Pr(C)}$ is equal to what you obtained with a two-way table.
