College Math
posted by Sigmund on .
A bucket contains orange tennis balls and yellow tennis balls from which 5 balls are selected at random, but assume that the bucket contains 7 orange balls and 6 yellow balls.
(1) What is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow?
(2) What is the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow?

(1) From 1, subtract the probability that none are yellow and the probability that none are orange. Every other possibility has one or more of both.
All orange probability = (7/13)(6/12)(5/11)(4/10)(3/9) = 0.0163
All yellow probability = (6/13)(5/12)(4/11)(3/10)(2/9) = 0.0047
10.01630.0047 = 0.979
(2) See what you can do using a similiar method. 
Thanks! I am still working on the second part.. and having a difficult time... but I am sure I will eventually figure it out. :)

From one, subtract the probabilities of:
(1) zero or one orange, (2) zero or two yellow and (3) one orange and one yellow.
Zero orange (all yellow):0.0047
Zero yellow (all orange):0.0163
One orange (4 yellow)): 5x(6/13)(5/12)(4/11)(3/10)(7/9)= 0.08159
One yellow (4 orange): 5*(7/13)(6/12)(5/11)(4/10)(6/9)= 0.1632
10.00470.01630.08160.1632 = 0.8974