m0x
I love juice too
Premium
join:2002-11-04
San Francisco, CA
| reply to Spider Man
Re: Math help please :)
Now, let's say I take 5 books randomly, without putting them back. What are the chances that I have at most 2 19$ books in the 5 I took?
Well, my chances of of getting a 19$ in one pick is (210/350) = 0.6
So my probability of getting a 14$ one is 1-0.6 = 0.4
There are 6 different outcome. I can get:
0 19$ books 5 14$ books
1 19$ books 4 14$ books
2 19$ books 3 14$ books
3 19$ books 2 14$ books
4 19$ books 1 14$ books
5 19$ books 0 14$ books
So I have to add up the probability of getting a 0, 1 and 2 books together to get at most 2 19$ books.
The formula for the probability is:
p(x) = ( 5! / (x!*((5-x)!)) ) * 0.6^x * 0.4^(5-x)
where x is the number or 19$ book you'll get.
So we'll add up p(0), p(1) and p(2)
That'll make: p(0) + p(1) + p(2) = 0.01024 + 0.0768 + 0.2304
The probability of getting at most 2 19$ books randomly from 5 books taken from your store is thus: 31.744%
Please accept my apology. People already answered the question and I've been studying non-stop the past days on my stats class. This is a very simple example. Stats are cool, go in science kids.
Oh, and I probably made a mistake or two.
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