• Feb 2nd 2010, 12:04 PM
bigroo
There are 10 different hats, what is the probability that if I got 4 of them that I would get at least 2 the same?

Would it be \$\displaystyle 1/10 * 1/9 * 1/8 * 1/7\$?

• Feb 2nd 2010, 12:44 PM
How many types of hat are there? Post the ENTIRE question.
• Feb 2nd 2010, 01:03 PM
bigroo
10 types of hat which are equally likely to be got.
• Feb 2nd 2010, 01:06 PM
Ok but how many hats are there in total? If there's only ten then you can't get two the same.
• Feb 2nd 2010, 01:10 PM
bigroo
Ok I see now, 10 different types of hat but presumably infinite hats in total. The question is ' if you get four attempts you will obtain at least 2 of the same type'.
You only have 4 attempts to choose.
• Feb 2nd 2010, 01:28 PM
Maybe this is wrong but I think it's...

1-Probability(all hats are different)

So you get your first hat with prob 1.
The prob that the second one is different is 9/10
The prob that the third one is different is 8/10
The prob that the fourth one is different is 7/10

So 1-Probability(all hats are different) = 1-(9/10*8/10*7/10) = 0.496.
• Feb 2nd 2010, 02:22 PM
Plato
Quote:

Originally Posted by bigroo
Ok I see now, 10 different types of hat but presumably infinite hats in total. The question is ' if you get four attempts you will obtain at least 2 of the same type'.
You only have 4 attempts to choose.

There are \$\displaystyle \binom{4+10-1}{4}\$ total number of ways to choose four hats.
There are \$\displaystyle \binom{10}{4}\$ ways to choose four hats all of different types.
• Feb 2nd 2010, 10:07 PM
bigroo
Quote:

Maybe this is wrong but I think it's...

1-Probability(all hats are different)

So you get your first hat with prob 1.
The prob that the second one is different is 9/10
The prob that the third one is different is 8/10
The prob that the fourth one is different is 7/10

So 1-Probability(all hats are different) = 1-(9/10*8/10*7/10) = 0.496.

So my answer was nearly the same as yours but instead of \$\displaystyle 1/10\$ you have 1.
• Feb 3rd 2010, 03:42 AM