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.
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: .$\displaystyle _{50}C_{23}$ possible outcomes.
There are 6 black balls. We want 6 of them.
. . There is: .$\displaystyle _6C_6\:=\:1$ way.
There are 44 red balls, and we want 17 of them.
. . There are: .$\displaystyle _{44}C_{17}$ ways.
Hence, there are: .$\displaystyle (1)(_{44}C_{17}) \:=\:_{44}C
_{17}$ ways to draw 6 black balls.
Therefore: .$\displaystyle P(\text{6 black balls}) \;=\;\frac{_{44}C_{17}}{_{50}C_{23}} \;=\;\frac{44!}{17!\,27!}\cdot\frac{23!\,27!}{50!}$
I think it comes out to: .$\displaystyle \frac{4807}{756,\!700} $