Posted by **Wing** on Tuesday, August 5, 2008 at 8:38pm.

A golf bag contains 6 white balls and 8 yellow balls. What is the probability of each event?

A) Drawing 3 white balls?

B.) Drawing 1 Yellow ball and 1 white ball?

Thanks

- MATH 12 -
**Reiny**, Tuesday, August 5, 2008 at 9:33pm
in A, are you drawing only 3 balls?

then prob = (6/14)(5/13)4/12) = 5/91

or C(6,3)/C(14,3) = 20/364 = 5/91

in B, are you drawing only 2 balls?

then (8/14)(7/13) = 4/13

or C(8,2)/C(14,2) = 28/91 = 4/13

- MATH 12 -
**Damon**, Tuesday, August 5, 2008 at 9:42pm
binomial distribution

p white = 6/14 = 3/7

p yellow = 1 - pwhite = 4/7

in three trials, what is the chance of drawing exactly 3 white?

The binomial coef of 3 out of three is 1

because 3!/(3!(3-3)!) = 1

P(3/3) = 1 * (3/7)^3 * (4/7)^0

so (3/7)^3

What is the probability of drawing exactly one white out of 2 trials?

C(2/1) = 2!/[1! (2-1)!] = 2

P(1/2) = 2 * (3/7)^1 * (4/7)^1

= 24/49

- MATH 12 -
**Damon**, Wednesday, August 6, 2008 at 9:25am
Use what Reiny wrote. I ignored the changes in probability as the bag contained fewer balls.

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