# Hypergeometric

• Mar 22nd 2011, 11:22 AM
dwsmith
Hypergeometric
From a box containing 5 white balls and 4 reds, two are randomly selected w/o replacement. Find

(a) Exactly 1 is white.
X number of white selected.
$P(X=1), \ N=9, \ k=5, \ n=2$

$\displaystyle p(1)=\frac{\binom{5}{1}\binom{4}{1}}{\binom{9}{2}} =\frac{5}{9}$

The book says the answer is 0.14. I don't know what I did wrong since what I have done seems correct.
• Mar 22nd 2011, 11:53 AM
dwsmith
I also have a question in regards to part (c).

Two white balls are selected, given at least one white is selected.
$\displaystyle P(2W|\text{at least} \ 1W)=\frac{P(2W\cap \text{at least} \ 1W)}{\sum_{x=1}^2\frac{\binom{5}{x}\binom{4}{2-x}}{\binom{9}{2}}=\frac{5}{6}}$

I don't know how to determine that intersection.
• Mar 22nd 2011, 12:23 PM
Plato
Question: what does "w/replacement" mean?
Is it with replacement? Or without replacement?
• Mar 22nd 2011, 12:26 PM
dwsmith
Quote:

Originally Posted by Plato
Question: what does "w/replacement" mean?
Is it with replacement? Or without replacement?

I forgot the o. It shold be w/o=without
• Mar 22nd 2011, 12:46 PM
Plato
$P(\text{2W}\cap\text{at least one W})=P(\text{2W})$