Population proportion

Question

A type of disease that affects people over the age of 60 is far more prevalent in men (2%) than in women (0.1%). There is a screening test used on people over 60 that has different accuracy rates for men and women.

What is the probability a person will test positive?

The prior probability for men to have a disease is 2 in 100 while for women it is 1 in 1000

So to find the probability that a random person selected has a diseases do we add the probabilities i.e 0.02+0.001 = 0.021 OR do we do (2+1)/ (100+1000) ?

Answer 1

You need to be on the same "scale".

men 2/100 = 20/1000

women 1/1000 = 1/1000 (duh)

Now you see the two ways you have formulated it are the same.

Answer 2

The probability of the disease (D) under the condition that the person is male (M) is 0.02 and under the condition that the person is female (F) it is 0.001: $$ \begin{align} p(D|M) &= 0.02,\\ p(D|F) &= 0.001. \end{align} $$ The probability of a random person having the disease is: $$ \begin{align} p(D) &= p(D|M) \cdot p(M) + p(D|F) \cdot p(F)\\ &= 0.02 \cdot p(M) + 0.001 \cdot p(F). \end{align} $$ As you can see, you need to know the probability of a person being female or male, otherwise, you cannot give an answer. Presuming that half of the people you consider are female, the result would be: $$ \begin{align} p(D) &= 0.02 \cdot 0.5 + 0.001 \cdot 0.5\\ &= 0.0105. \end{align} $$