statistics
posted by kp .
A jar contains 7 red balls, 5 green balls, 4 blue balls, and 3 white balls. A sample of size 7 balls is selected at random without replacement . Find t the probability that sample contains two red balls, 2 green balls, 2 blue balls, and 1 white ball.

Let's simplify the problem to having
7 red balls,
5 green balls, and
4 blue balls
(for a total of 16).
Find the probability of drawing
2 red balls, 2R
2 green balls, 2G
2 blue balls, 2B
(total of 6 balls), then
Below we will use the notation of
(n,r)=combination (n,r)
=n!/((nr)!r!)
Number of ways of drawing
2R,2G and 2B would be
for 2R: (7,2)=21
for 2g: (5,2)=10
for 2B: (4,2)=6
and number of ways of drawing 6 balls out of 16: (16,6)=8008
So probability
P(2R+2G+2B)=(7,2)(5,2)(4,2)/(16,6)
=21*10*6/8008
=45/286
Extend the idea to the problem in hand, and post for answer checking if you wish.
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