8 balls are in a box. Each ball can be either blue or red. If there are d red balls in the box, calculate the probability p(d) that two randomly selected balls from the ball are both blue. Sketch p(d).

p(d) = (8-d)/8 * (7-d)/7 = (d^2 - 15d + 56)/56

Now suppose that the box contains 8 balls whereof d red balls and 8-d blue balls. Let Z be the number of red balls when two balls are selected randomly from a box. Find the probability function for Z, end determine E(Z) and Var(Z).


\(P(Z) = \frac{_{8-d}C_{2-Z}*_{d}C_{Z}}{_{8}C_{2}} = P(Z) = \frac{_{8-d}C_{2-Z}*_{d}C_{Z}}{28}\)

Is my thinking correct? Is there an easier way to write this?

 

n(A) = 2 balls, not blue = (8-d) choose 2 = (8-d)!/(8-d-2)! 2!
n(S) = 8 choose 2 = 8!/(8-2)! 2!

so probability of choosing 2 blue balls is n(A)/n(S)
= (8-d)! 6! /(6-d)! 8! = (8-d)!/(6-d)! * 56