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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?

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  • $\begingroup$ @AlexR I am not so sure about that. See my answer. $\endgroup$ – drhab Jun 16 '15 at 10:53
  • $\begingroup$ @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. $\endgroup$ – AlexR Jun 16 '15 at 11:08
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Your answer on the second question is based on the interpretation: "If the pc is not infected by mail, then what is the probability that it is infected?"

If the second question is asked in the knowledge that the pc is infected then I think you should go for the following interpretation: "What is the probability that the infected pc is not infected by mail?"

To be found is then: $$P\left\{ M^{c}\mid M\cup B\right\} $$

$$P\left\{ M^{c}\mid M\cup B\right\} =\frac{P\left(B-M\right)}{P\left(M\cup C\right)}=\frac{0.15}{0.55}=\frac{3}{11}$$

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  • $\begingroup$ Thanks! I guess I misinterpreted the question. $\endgroup$ – Seneca Jun 16 '15 at 11:15

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