1. ## Probability Question

Suppose we have box containing 50 balls of which 6 are black and the rest red.
Without seeing the balls we take 23 balls out of the box. My question is what is the probability that 6 of these 23 balls are black.
Thank you.

2. Originally Posted by ciccio
Suppose we have box containing 50 balls of which 6 are black and the rest red.
Without seeing the balls we take 23 balls out of the box. My question is what is the probability that 6 of these 23 balls are black.
Thank you.
Read this: Hypergeometric distribution - Wikipedia, the free encyclopedia

3. thanks. I almost got it, need to read it again ,

4. Hello, ciccio!

We have a box containing 50 balls of which 6 are black and the rest red.
Without seeing the balls we take 23 balls out of the box.
What is the probability that 6 of these 23 balls are black?

There are 50 balls. We take 23 of them.
. . There are: . $_{50}C_{23}$ possible outcomes.

There are 6 black balls. We want 6 of them.
. . There is: . $_6C_6\:=\:1$ way.
There are 44 red balls, and we want 17 of them.
. . There are: . $_{44}C_{17}$ ways.

Hence, there are: . $(1)(_{44}C_{17}) \:=\:_{44}C
_{17}$
ways to draw 6 black balls.

Therefore: . $P(\text{6 black balls}) \;=\;\frac{_{44}C_{17}}{_{50}C_{23}} \;=\;\frac{44!}{17!\,27!}\cdot\frac{23!\,27!}{50!}$

I think it comes out to: . $\frac{4807}{756,\!700}$