Posted by **Sigmund** on Saturday, February 28, 2009 at 11:48pm.

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?

- College Math -
**drwls**, Sunday, March 1, 2009 at 12:05am
(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

1-0.0163-0.0047 = 0.979

(2) See what you can do using a similiar method.

- College Math -
**Sigmund**, Sunday, March 1, 2009 at 1:25am
Thanks! I am still working on the second part.. and having a difficult time... but I am sure I will eventually figure it out. :)

- College Math (2) -
**drwls**, Sunday, March 1, 2009 at 2:02am
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

1-0.0047-0.0163-0.0816-0.1632 = 0.8974

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