Probability defective computer problem

[snoggerT]

A computer retail store has 14 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail store or the buyer, 3 of the computers in stock have defective hard drives. Assume that the computers are selected at random.

What is the probability that exactly one of the computers will be defective?

What is the probability that at least one of the computers selected is defective?

The Attempt at a Solution

- I'm at a loss on this problem. I've been able to work all the other problems from the homework, but I just can't seem to figure this one out. Can someone please help?

 

[Gregg]

$$P(X=1) = 3\frac{3}{14}\frac{11}{13}\frac{10}{12}$$

I think that because there are 14 computers and 3 broken ones the odds of choosing a broken one at first will be 3 in 14, then to avoid it subsequently it will be 11 in 13 then 10 in 12. There are 3 equally likely scenarios in which this can happen.


At least 1 is P(X=1) + P(X=2) + P(X=3) or 1-P(X=0) which is easiest to compute.

$$P(X\ge 1) = 1-\frac{11}{14}\frac{10}{13}\frac{9}{12}$$

I think this OK.

 

[snoggerT]

$$P(X=1) = 3\frac{3}{14}\frac{11}{13}\frac{10}{12}$$

I think that because there are 14 computers and 3 broken ones the odds of choosing a broken one at first will be 3 in 14, then to avoid it subsequently it will be 11 in 13 then 10 in 12. There are 3 equally likely scenarios in which this can happen.


At least 1 is P(X=1) + P(X=2) + P(X=3) or 1-P(X=0) which is easiest to compute.

$$P(X\ge 1) = 1-\frac{11}{14}\frac{10}{13}\frac{9}{12}$$

I think this OK.

- The first part is right, but I'm not exactly sure on why yet. Can you maybe elaborate a little more?

 

[njama]

a)
$$P=\frac{C_{11}^2 * C_3^1}{C_{14}^3}$$

There are 11 working and 3 damaged computers, and a total of 14.

Its like combining 2 computers (from 11) that work and 1 computer (from 3) that do not work.

You will try something for b) after knowing what a) is