A computervirus has a probability of 0.4 to infect your pc through mail and 0.3 trough your browser. The probability that your pc is infected by both mail and browser is 0.15. $$P(M)=0.4$$ $$P(B)=0.3$$ $$P(M\cap B)=0.15$$
It seems they are not independent.
What is the probability you don't have a virus?
Is this: $$ =1-P(M\cup B)=1-P(M)-P(B)+P(M\cap B)=1-0.4-0.3+0.15=0.45 $$
What is the probability that you have a virus which doesn't originate from a mail?
I translated the above to: $$ P\left ( (M\cup B) \mid \overline{M} \right)= \frac{P\left ((M\cup B)\cap \overline{M} \right )}{P(\overline{M})} = \frac{P(B\cap \overline{M} )}{P(\overline{M})}= \frac{P\left ( B\setminus (B\cap M) \right )}{1-P(M)} = \frac {0.3-0.15}{1-0.4} = \frac {0.15} {0.6} = 0.25 $$
Am I correct?