Thanks, you got it.

Since I'm very weak with combinations and probability, I'm trying to solve it in the following ways.

Three of four ways (1,2,4) work, one doesn't (3).

Could anybody say what's wrong in my third approach?

Thanks!

1) probability approach:picking first black mice = \(4/10\) = \(2/5\)

picking second black mice = \(3/9\) = \(1/3\)

picking third black mice = \(2/8\) = \(1/4\)

picking three black mice = \(2/5*1/3*1/4=1/30\)

2) combinatorial approach:**Quote:**

Number of ways to choose 3 black mice = \(C(4,3)\) (Picking any three of the four)

Number of ways to choose any 3 mice = \(C(10,3)\) (Picking any three of the ten)

So probability that all three mice are black is \(4/120\) or \(1/30\)

3) reversal probability approach:The probability of picking three black mices equals to 1 minus the probability of picking three white mices plus the probability of picking two white mices and one black plus the probability of picking one white mices and two black.

That is:

P(b,b,b) = 1 - [P(w,w,w) + P(w,w,b) + P(w,b,b)]

P(w,w,w) = \(6/10*5/9*4/8=1/6\)

P(w,w,b) = \(6/10*5/9*4/8=1/6\)

P(w,b,b) = \(6/10*4/9*3/8=1/10\)

so:

P(b,b,b) = \(1- (1/6+1/6+1/10) = 1 - 13/30 = 17/30\)

4) reversal combinatorial approach:Same approach of the above one.

Number of ways to choose any 3 mice = \(C(10,3)\) = 120 (Picking any three of the ten)

Number of ways to choose one mice white and two mices black = \(C(6,1)\)*\(C(4,2)\) = 36

Number of ways to choose two white mices and one mice white = \(C(6,2)\)*\(C(4,1)\) = 60

Number of ways to choose three white mices = \(C(6,3)\) = 20

So probability that all three mice are black is:

\(1 - [C(6,1)*C(4,2)+C(6,2)*C(4,1)+C(6,3)]/C(10,3) = 1 - (36+60+20)/120 = 4/120 = 1/30\)