Reasoning gone wrong? While I was going through the previous year question paper of a college entrance examination conducted in our country, I came across this particular problem.

The probability of men getting a certain disease is $\frac{1}{2}$ and that of women getting the same disease is $\frac{1}{5}$. The blood test that identifies the disease gives the correct result with probability $4/5$. Suppose a person is chosen at random from a group of $30$ males and $20$ females, the blood test of the person is found to be positive. What is the probablity that the chosen person is a man?

Now at that time I wasn't aware of using conditional probability so I went a quite different way.
The test could have been correct or maybe wrong.
If we consider the case where the test is correct.
Number of diseased men $= 30(\frac{1}{2})=15$.
Number of diseased women $ = 20(\frac{1}{5})=4$.
Given that the test is positive the probability is
$(\frac{4}{5})(\frac{15}{15+4})$
If the test is incorrect then that particular person do not have the disease.
Total men and women without the disease $15+16=31$
Probabilty for this case :
$(\frac{1}{5})(\frac{15}{15+16})$
Adding both cases gives $\approx0.72$
However the real ansewer is around $0.70$ which is obatined by using conditional probabilty.
After this I did dug into conditional probabilities but I am still curious about why my method didn't worked.
If you could point it out, I will be very grateful.
Thanks
 A: I try to follow your approach (without explicit use of conditional probabilities) and give you the correct answer.
The test will be positive for $4/5$ of the total persons with the desease i.e. $\color{blue}{15}$ males and $\color{red}{4}$ females over $\color{blue}{30}+\color{red}{20}=50$, or for $1/5$ of the total persons without the desease i.e. $\color{blue}{15}$ males and $\color{red}{16}$ females over $50$. Hence the desired probability is
$$\frac{\frac{4}{5}\cdot\frac{\color{blue}{15}}{50}+\frac{1}{5}\cdot\frac{\color{blue}{15}}{50}}{\frac{4}{5}\cdot\frac{\color{blue}{15}+\color{red}{4}}{50}+\frac{1}{5}\cdot\frac{\color{blue}{15}+\color{red}{16}}{50}}=\frac{75}{107}\approx 0.7009.$$
A: The fact that the probability of a man getting the disease is $\frac{1}{2}$, doesn't mean that exactly $\frac{1}{2}$ of a population of men will get the disease. You can't know beforehand how many men will have the disease. You can only calculate the probability using the Bernoulli Distribution: The probability than $n$ men out of $30$ will get the disease is $$ P(X = n) = {30 \choose n} p^{n} (1-p)^{30 - n}, $$ where $ p = \frac{1}{2}$. The same logic follows for women. The probability of a person getting the disease applies only for that specific person, not for an entire population.
