If you have 50 balls and 10 are defective, what is the probability that when you pick 30 none will be bad?

Question

Please explain procedure.

Answer 1

I would say:

40!*20! / 10! / 50!


The first attempt of picking... your chance is 40 out of 50, right?
40/50

The second attempt, one has already been taken out... but all the defective ones are still there. So, you have 39 possible right choices out of a total of 49 possible choices.
39/49

So, youre total chances after only two attempts is
40/50 * 39/49

This continues for a total of 30 terms

40*39*38*37.... *11 / 50*49*48*47... *21

Or, more simply written:
40!/10! / 50!/20! =
(40! * 20!) / (10!* 50!)

Answer 2

If you draw one you have a 40 in 50 chance of getting a good ball.
If you draw 2 you would have a 39 in 49 chance that none are bad

etc.
so it would be 40*39*38*...*11/(50*49*....21)
So the 40...21 would cancel each other out and you would have:
20*19*18*17*16*15*14*13*12*11/50*49*48*47*46*45*44*43*42*41

which would be approximately 1 in 55600 chance that none would be bad.

Answer 3

You have 10 defective balls and 40 balls that are not defective.

Probability = (40C30) / (50C30)

= [40!/(30!*10!)] / [50!/(30!*20!)]

= (40!*20!) / (50!*10!)