Q:

H1N1 Probability Problem

Suppose that the probability of being affected by H1N1 flu virus is
  0.02, the probability of those who regularly wash their hands among those affected by the H1N1 virus is known as 0.3, In general, if the
  probability of people who washing their hands regularly between the
  whole people (whether they affected or not by the virus) is 0.6, Find
  the possibility of those who regularly wash their hands to be affected
  by the virus.

Proposed Solution:
lets A be the event of people who affected with H1N1 virus
and  B be the event of people who regularly washing their hands

In this case 
P(A) = 0.02
P(A and B) = 0.3
P(B) = 0.6
P(A|B) = ?

Rule:
P(A|B) = P(A and B) / P(B)
       =       0.3 / 0.6 = 0.5

Well, my question is:
Are my assumptions and the proposed solution right or not? if not please correct my answer.

Thanks in advance.
:)

A:

If I correctly understood your problem, you should use Bayes formula:
$$
P(infect|wash) = \frac{P(wash|infect)P(infect)}{P(wash)}
$$
where $P(infect) = 0.02, P(wash)=0.6, P(wash|infect) = 0.3$.

