But, the probabilities of those 4-draws are not all the same.

bobbym

2013-09-06 10:38:43

Hi all;

That is all the ways to draw 4 coins from 5 and 3. There are 15 total ways and four winners. So I will have to go with 11 / 15 jist as before. Also, the OP already verified this.

anonimnystefy

2013-09-06 10:19:35

I think he might be right. I am getting 13/14.

Code:

(Select[Permutations[{p, p, p, p, p, n, n, n}], Count[Take[#, 4], n] < 3 &] // Length)/(Permutations[{p, p, p, p, p, n, n, n}] // Length)

Anynomous

2013-09-06 09:45:44

This solution is not right, i tried it. Can you show how you got it, or something

bobbym

2013-01-23 15:45:31

Hi;

cooljackiec

2013-01-23 14:48:49

I have a bag with 5 pennies and 3 nickels. I draw coins out one at a time at random. What is the probability that I haven't removed all 3 nickels after 4 draws?