Probability of balls in a box

[deadguybob51]

I think I got this one, but I'm not sure, I never know if I counted everything up right.

A box contains 30 red balls, 30 white balls, and 30 blue balls. If 10 balls are selected at random, without replacement, what is the probability that at least one color will be missing from the selection?

Here's what I did:
P(at least 1 color missing) = 3*P(all 1 color)+3*P(2 colors)
P(all 1 color) = [sub:10e0qrs9]30[/sub:10e0qrs9]C[sub:10e0qrs9]10[/sub:10e0qrs9]/[sub:10e0qrs9]90[/sub:10e0qrs9]C[sub:10e0qrs9]10[/sub:10e0qrs9] = 30!80!/20!90!
P(2 colors) = [sub:10e0qrs9]60[/sub:10e0qrs9]C[sub:10e0qrs9]10[/sub:10e0qrs9]/[sub:10e0qrs9]90[/sub:10e0qrs9]C[sub:10e0qrs9]10[/sub:10e0qrs9] = 60!80!/50!90!

P(at least 1 color missing) = 3*(30!80!/20!90! + 60!80!/50!90!)

I'm unsure if I included all the possibilities in my first step, and just a bit more confident that everything got counted up right. Thanks.

 

[soroban]

Hello, deadguybob51!

I think I have an approach to this one . . .

A box contains 30 red balls, 30 white balls, and 30 blue balls.
If 10 balls are selected at random, without replacement, what is the probability
that at least one color will be missing from the selection?

\(\displaystyle \text{There are: }\:{90\choose10}\text{ possible ways to draw ten balls.}\)


\(\displaystyle \text{"At least one color missing" means the 10 balls are of [1] one color or [2] two colors.}\)

\(\displaystyle \text{[1] The ten balls are of one color.}\)
. . \(\displaystyle \text{There are 3 choices of color.}\)
. . \(\displaystyle \text{There are }\,{30\choose10}\text{ ways get three of that color.}\)
. . \(\displaystyle \text{Hence, there are: }\:3\!\cdot\!{30\choose10}\text{ ways to get three of one color.}\)


\(\displaystyle \text{[2] The ten balls are of two colors.}\)
. . \(\displaystyle \text{There are 3 choices for the two colors.}\)
. . \(\displaystyle \text{There are: }\,{60\choose10}\text{ ways to get ten balls from those sixty balls.}\)
. . \(\displaystyle \text{But }\,2\!\cdot\!{30\choose10}\text{ of those ways are of one color.}\)
\(\displaystyle \text{Hence, there are: }\,3{60\choose10} - 2{30\choose10}\text{ ways to get exactly two colors.}\)


\(\displaystyle \text{Therefore, there are: }\,3{30\choose10} + 3{60\choose10} - 2{30\choose10}\text{ to get ten balls with at least one missing color.}\)

\(\displaystyle \text{Now divide by }\,{90\choose10}\)