Pick candies (A probability problem)

問題來源: haha2000 於 06-10-24 13:56:08
Given a jar, there are 10 red candies, 20 blue candies, 30 green candies, what is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken all the red candies out?

Deuss:  7/12

There are totally 60! Permutations.

Now count those end with RG* or RB*, where * means 0 or more occurrences of the preceding color.

In general, assume m+n+1 numbered balls, with m of color X, 1 of color Y, and n of color non-X and non-Y, the number of permutations ending with YX* is
C(m+n+1,n+1)m!n!. This is because, there are m!n! ways of permuting m X’s and n non-X non-Y separately. On the other hand, for each (n+1) picks out of (m+n+1), treat the last of the (n+1) as Y, and the rest n as non-X non-Y. Then treat the unpicked m as X. Apply the permutation to the picking, we get a permutation ending with YX*. Vice versa.

If there are 10 variations of color Y, then multiply the above by 10. From this we know, the number of RB* ending permutations is 10C(60,20)39!20!, and the number of RG* is 10C(60,30)29!30!. Since they both count the case of R ending, need to minus 59!10.
The total number of RB* or RG* ending is
10C(60,20)39!20!+ 10C(60,30)29!30!- 59!10
=10(60!/40+60!/30-59!)
The probability of RB* or RG* is 10(1/40+1/30-1/60)=5/12
The probability of R{..B..G..|..G..B..}+ is then 1-5/12=7/12