Conditional probability

Quote:

Originally Posted by lost1

Basicaly the Q is:
30% use Mac, 50% use Windows, 20% use Linux.
Also 60% of mac has virus, 80% of Windows has virus, 10% Linux has virus.
A random computer is chosen and is found to have virus. What is the probability that this computer is a Windows PC?

To a similar question my tutor recomended drawing up a table so I did that for this question as well but cant get the right answer.

| Virus | Non-Virus | TOTAL
Mac | 0.3*0.6= 0.18 | 0.3-0.18=0.12 | 0.3
Windows | 0.5*0.8= 0.4 | 0.5-0.4 =0.1 | 0.5
Linux | 0.2*0.1= 0.02 | 0.2-0.02 =0.18| 0.2
TOTAL | 0.6 | 0.4 | 1

I would have thought the answer is 0.4 but its 2/3 =S
Can someone please point out where iv gone wrong?

Thanks

Sorry didnt know how to make a table and the extra spaces i tried to use dont seem to work...but you get the idea
thanks

Suppose there are 100 computers, 30 Macs, 50 Win, 20 Linux.

Of the Macs 16 have v.
Of the Win 40 have v.
Of the Lin 2 have v.

There are a total 58 with v. and of those with v. 16/58 are macs, 40/58 are Win and 2/58 are Lin.

So the probability that a computer with v. is Win is 40/58.

CB