Posted by **John** on Saturday, September 26, 2009 at 7:11pm.

1. In a large box are 20 ping-pong ballsâ€”10 of them are red and numbered 1 through 10, and the other 10 of them are green and also numbered 1 through 10. Two balls are selected from the box without replacement.

1.

a.Find the probability that the first ball selected is red and even.

2.

b.Find the probability that the first ball selected is red or even.

3.

c.Find the probability that both balls selected are green.

4.

d.Find the probability that the first ball selected is green or the second ball selected is red.

5.

e.Find the probability that at least one of the balls is odd.

- statistics -
**economyst**, Sunday, September 27, 2009 at 10:26pm
There are 20 balls, this is the denominator for a and b

a) there are 5 red even balls, prob=5/20

b) there are 15 red or even balls, prob=15/20

c) there are 20-choose-2 ways to choose 2 balls = 20!/2!*(20-2)! = 19*20/2 ways. This will be the denominator in c

Number of ways to choose 2 green are 10-choose-2. This is the numerator.

d) probability that the first is red AND the second is green is P=(10/20)*(10/19). So d is (1-P)

e) Probability both are even is P=(10/20)*(9/19). so, e is (1-P)

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