computer virus prob.: If program is selected at random...

A computer store has 10 copies of a word processing program, 12 copies of a spread sheet program, and 8 copies of a drawing program. Three of the word processing, four of the spread sheet and two of the draw programs are infected with a computer virus. If a program is selected at random, the probability it is infected with a virus is. Ok this is how I broke it down:

My answer is 7/30 chances it will have a virus. Is this right? If not, please show me how to set this one up. Thanks so much for taking time to look over this problem.

We all know where the 30 comes from.
But where did you get the 7 from? Is that a guess?

not really a guess

Just trying to understand how to set this one up. Ok I see it's 11/30, I guess I'm thinking how many people will pull one and buy it. So 11 people could end up with a computer virus.

No! There are 9, 3+4+2, infected. So the answer is 9/30.

I don't have that as a possible answer.

1. 11/30
2. 7/30
3. 3/10
4. 1/4 Is all I have as possible answers. I know 9/30 works but I don't have that on the list. So it looks like 3/10 could work also right. thanks for helping me with this question again.

\(\displaystyle \frac{9}{{30}} = \frac{{\left(\not 3 \right)\left(3 \right)}}{{\left(\not 3 \right)\left( {10} \right)}} = \frac{3}{{10}}\)
Man, you have got to think about what all this means.

thank you

I'm trying...I'm trying trust me. I see what you did and what I thought it would be was right after trying to work it out. Thanks for being a big help.