Probability word problem.?

Question

A department store receives a shipment of 23 new portable radios. There are 5 defective radios in the shipment. If 4 radios are selected for display, what is the probability that 2 of them are defective?

Answer 1

4 will be selected, of those two will be defective (out of 5 possible defectives) and two will NOT be defective (out of 18 non-defectives).

The number of ways to select the two defective is 5C2=5!/(2!3!)

The number of ways to select the two nondefective is 18C2=18!/(2!16!).

The total number of ways to select 4 radios out of 23 is 23C4=23!/(4!19!)

Multiply and divide to get the probability:
(5C2)(18C2) / (23C4)

The answer comes to approximately .17278

Answer 2

I also don't want to do your homework for you -- but let me put you on a better track than the clowns who already answered.

You have to figure out how many ways *combinations) there are to pick two good radios and two bad ones & divide that by the number of combinations.

So -- you have to find out how many ways you can pick two bad radios -- that is the number of combinations using 5 things & pulling two (5!/(2!*3!))

You need to multiply that by the number of ways you can pull two of the good radios. That is the number of combinations of the 18 things pulled 2 at a time (18!/(2!*16!))

Divide this by all the possible combinations of 23 things taken four at a time.

Answer 3

I don't want to do your homework for you, so I'm just going to put you on the right track and let you figure it out from there.

If 5 out of 23 radios are defective, then _____ out of 4 radios will be defective.

There are four possible ways to fill in the blank: 1, 2, 3, or 4.
One out of those 4 answers is 2.

Good luck!

Answer 4

You should be able to find this problem out on your own... that is if you've been listening to your teachers. Just think about it for a bit, and next time, listen to your teachers.