There are 12 balls in a box, among which 3 are red, 4 are blue, and 5 are green.

Question

There are 12 balls in a box, among which 3 are red, 4 are blue, and 5 are green. Consider a random sample of size 6. What is the probability that there are at least 1 red, 1 blue, and 1 green balls in the sample?

I am starting with the complement which is the probability that none are red, blue and green then subtracting it from 1.

1 -. ((9 choose 6) + 8 choose 6) + (5 choose 6 )) / (12 choose 6)

I feel like I am somehow overcounting and need to subract something from this. Can someone expain how I cam overcounting and provide correct solution?  

Explanation / Answer

Total number of balls: 3+4+5 = 12

Number of ways of selecting 6 balls out of 12 is C(12, 6) =924.

Since each color has less than 6 balls so it is not possible to select sample of single color.

Now number of ways of seelcting only red and blue balls is C(7,6)

Number of ways of seelcting only red and green balls is C(8,6)

Number of ways of seelcting only blue and green balls is C(9,6)

So number of ways of selecting ball without at least one color is  C(7,6)+ C(8,6)+ C(9,6) = 7+28+84=119

So number of ways of seelcting 6 balls with all three colors is 924-119 = 805

Hence, required probability is 805 / 924 = 0.8712

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