[Guest]

This question seems so simple it is frustrating.

Suppose you plan to purchase stock in 3 electrical utility companies from among 7. Unknown to you, 2 of the companies will experience serious difficulties. If you randomly select the three companies from among the 7, what is the probability that you select:

1) None of the companies with difficulties?

2) One of the companies with difficulties?

3) Two of the companies with difficulties?

Thank you very much in advance.

Edited October 10, 2011 by Bollinger

[curr3nt]

Start with total number of selections.

Since order of the companies selected doesn't matter it is a Combination, not a Premutation. So from 7 choose 3 using n!/(r!(n-r)!). 7!/(3! * (7-3)!) = (7*6*5)/(3*2*1) = 35 different combinations.

You could go the long way and write out all 35 combinations and count each that have the bad companies. Or there is another way...

edit -

I am assuming that you can't pick the same company more than once.

Edited October 10, 2011 by curr3nt

[curr3nt]

1) Let's say companies #6 and #7 are the bad ones. That leaves 5 good companies. So you want to pick 3 from the 5 good companies.

2) You picked one bad company. First let us say you picked #6. You have 2 more to choose from the 5 good companies. Repeat for #7.

3) You chose #6 and #7 and now you need to chose one more.