1. ## Probabilité

Hi,

a bag contains n balls, among them there has one, and another appointee named b. It takes balls p bag. What is the number of draws that contain a and b.

A bag contains n balls, among them there has one appointed a and another appointed b. We take p balls from the bag. What is the number of draw that contains a and b? and the number of draw that contains just a and not b???

2. Originally Posted by lehder
Hi,

a bag contains n balls, among them there has one, and another appointee named b. It takes balls p bag. What is the number of draws that contain a and b.

A bag contains n balls, among them there has one appointed a and another appointed b. We take p balls from the bag. What is the number of draw that contains a and b? and the number of draw that contains just a and not b???
Hi lehder,

if "a" and "b" are in the selection of p balls, chosen from n,
then there remain p-2 balls to choose from the remaining n-2.

Hence there are $\binom{n-2}{p-2}$ selections containing "a", "b" and p-2 other balls.

If the p balls contain "a" but not "b",

then we have n-2 balls to choose from and we require p-1 new additional balls to join with "a".

Hence there are $\binom{n-2}{p-1}$ selections of "a" with p-1 others, none of which are "b".