# Thread: Probability problem gone sour

1. ## Probability problem gone sour

From 27 pieces of luggage, an airline luggage handler damages a random sample of four. Suppose 8 of the 27 are insured. Calculate the probability that exactly two of the four damaged pieces are insured.

Let's hold off with the solution to that problem and consider the following problem.

Suppose there are 27 pieces of luggage of which 8 are insured. Randomly pick 4 luggage and find P(exactly 2 of the 4 are insured). My answer, using hypergeometric, is 8C2*19C2/27C4

Now back to the original question. The solution manual states that p(that exactly two of the four damaged pieces are insured) = 8C2*19C2/27C4. But it is not (in my opinion) taking into account that 2 insured is coming from 4 damaged.

Can someone please enlighten me on this?

Thank you,

Steven

2. ## Re: Probability problem gone sour

The solution is correct.

Pick 4 pieces of luggage to be damaged. There are $27C4$ ways to do this.

How many ways can 2 of those 4 damaged pieces be insured. $8C2 \times 19C2$

Divide. Just as you've done.

Step 3 builds the 4 damaged pieces by selecting two insured and 2 not so you know the 2 insured make up 2 of the damaged pieces.

3. ## Re: Probability problem gone sour

Originally Posted by romsek
The solution is correct.

Pick 4 pieces of luggage to be damaged. There are $27C4$ ways to do this.

How many ways can 2 of those 4 damaged pieces be insured. $8C2 \times 19C2$

Divide. Just as you've done.

Step 3 builds the 4 damaged pieces by selecting two insured and 2 not so you know the 2 insured make up 2 of the damaged pieces.
Oh, OK. I see it now. Thanks!