Question

# A box contains 10 red balls, 10 white balls, and 10 blue balls. Five balls are...

A box contains 10 red balls, 10 white balls, and 10 blue balls. Five balls are selected at random, without replacement. Let X be the number of colors will be missing from the selection. Determine the probability mass function of X.

Total number of balls: 10+10+10=30

Number of ways of selecting 5 balls out of 30 is C(30,5).

At most 2 colors can be missing from the selection so X can take values 0, 1 and 2.

When X=1:

That is 1 color is missing. It means that sample has 2 colors. That is sample has 2 colors: red and white or red and blue or white and blue

Number of ways of selecting 5 balls from 10 red and 10 white balls:

Number of ways of selecting 5 balls out of 20 (red and white balls) is C(20,5). Number of ways of selecting 5 balls of only white colors C(10,5). Number of ways of selecting 5 balls of only red colors C(10,5). We need to find the number of ways of selecting at least one red and at least one white ball is

C(20,5) -2* C(10,5) = 15000

Likewise number of ways of selecting 5 balls from 10 red and 10 blue balls: 15000

Likewise number of ways of selecting 5 balls from 10 white and 10 blue balls: 15000

That is number of ways of selecting exactly 2 colors is: 15000 + 15000 +15000 = 45000

So,

P(X=1) = 45000 / C(30,5) = 0.3158

When X=2:

That is 2 colors are missing. It means that sample has 1 color. That is sample has 1 color: red or white or blue

Number of ways of selecting 5 balls from 10 red: C(10,5) = 252

Number of ways of selecting 5 balls from 10 white: C(10,5) = 252

Number of ways of selecting 5 balls from 10 blue: C(10,5) = 252

That is number of ways of selecting exactly 1 color is: 252+252+252= 756

So,

P(X=2) = 756 / C(30,5) = 0.0053

Now,

P(X=0) = 1 - P(X=1) -P(X=2) = 1 - 0.3158 - 0.0053 = 0.6789

Following table shows the pdf:

 X P(X=x) 0 0.6789 1 0.3158 2 0.0053

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