w****0
发帖数: 803 | 1 You are given 10 empty boxes. You distribute 5 black balls to these empty
boxes so that no two black balls can go into the same box. Now you have 4
white balls and you distribute them in the same fashion but this time a
white ball can go into a box where you put a black ball. That is, in the end
, you might have at least 5 full boxes or at most 9 full boxes. What is the
probability of choosing an empty box? | T*****u
发帖数: 7103 | | T*****u
发帖数: 7103 | | T*****u
发帖数: 7103 | | c***6
发帖数: 61 | 5 total = C54*C50 + C53*C51 + C52*C52 + C51*C53 + C50*C54
5 4 3 2
1 EMPTY BOX
if the prob of choosing each box is equal
total[5]/total * 5/10 + total[4]/total * 4/10 + total[3]/total *3/10 + total
[2]/total * 2/10 + total[1]/total *1/10
right? | p***0
发帖数: 233 | 6 Consider this problem in two parts
First part:what's the probability of forming 5 full boxes, 6 full boxes, etc
. Denote as p_1i, where i is [5,6,7,8,9] for each case.
Second part: for each case mentioned in the first part, what's the
probability of choosing the empty one. Denote as p_2i.
The second part is straightforward.
5 full boxes: 5/10
6 full boxes: 4/10
7 full boxes: 3/10
8 full boxes: 2/10
9 full boxes: 1/10
To solve the first part:
It's easy to find that the position of black balls don't involve any
probability. So we only care about how the white balls are put.
Denote C(k,n) as the k-combinations in n elements:
probability of generating 5 full boxes: C(4,5)/C(4,10)
probability of generating 6 full boxes: C(1,5)*C(3,5)/C(4,10)
probability of generating 7 full boxes: C(2,5)*C(2,5)/C(4,10)
probability of generating 8 full boxes: C(3,5)*C(1,5)/C(4,10)
probability of generating 9 full boxes: C(4,5)/C(4,10)
The final result is the sum (p_1i*p_2i) | E****t
发帖数: 6 | |
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