Cr and Probability

[peterliang]

Given 5 defective light bulb, and 15 good ones. Find probability of picking 3, and getting exactly 2 defective.

I thought it wud be

(5 pr 2) x 15 / (20 pr 3)

but its really cr,

can someone explain this to me? because i thought order would matter, because you can get 2 of the same defective bulb two different ways ,and so forth.

[AuSmith]

Usually with these probability questions, you can do either combinations or permutations. Say, out of n balls, k of them are red. If you draw r balls, the probability of all of them being red is

[unparseable or potentially dangerous latex formula]

Back to your problem, when you are counting the number of ways to draw 2 defective bulbs and 1 good bulb, it's as if you were choosing the defectives from one box and the goods from another, which is different from mixing them all together in one box. I'll explain.

This might be the setup in the background of your reasoning to find the number of ways of picking 2 defective bulbs, 1 good: [unparseable or potentially dangerous latex formula] is [unparseable or potentially dangerous latex formula], which comes from

(# ways to draw a defective bulb)(# ways to draw a defective bulb after the first defective is drawn)(# ways to draw a good bulb)

This setup leaves out the possibilities of drawing a good bulb on the first and second tries. That's why your answer was [unparseable or potentially dangerous latex formula] the size it needed to be. You could add in the other possibilities get the correct answer.

[unparseable or potentially dangerous latex formula]

And, [unparseable or potentially dangerous latex formula]

If you use the identity,

[unparseable or potentially dangerous latex formula]

you can get the correct answer you had.

Basically, you can work this problem with either permutations or combinations, but you are trying to mix them.

So, do you now thoroughly understand how

[peterliang]

ah! thanks i understand it now.
I guess most of the time it wud be easier to use Cr

[AuSmith]