# Conditional probability exercise infected by virus

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?

• @AlexR I am not so sure about that. See my answer. – drhab Jun 16 '15 at 10:53
• @drhab On second thought, your interpretation makes more sense to me (as in "What I would ask myself") The wording isn't very clear there. – AlexR Jun 16 '15 at 11:08

To be found is then: $$P\left\{ M^{c}\mid M\cup B\right\}$$
$$P\left\{ M^{c}\mid M\cup B\right\} =\frac{P\left(B-M\right)}{P\left(M\cup C\right)}=\frac{0.15}{0.55}=\frac{3}{11}$$