can not get the right answer (probability topic)

2007-01-28 6:07 am

A train consists of n cars, Each of m passengers (m>n) will choose a car at random to ride in. what is the probability that (a) there will be at least one passenger in each car;
(b) exactly r(r

更新1:

b)exactly r (r is less than n)cars remain unoccupied? the correct answer should be (a) (1/n^m)x[sum of (-1)^ix(n C i)(n-i)^m from i=0 to i=n] (b) (1/n^m)x(nCr)x[sum of (-1)^i x ( (n-r) C i ) x (n-r-i)^m from i=0 to i=n-r ],,how??what I think is (mCn) x n! x n^(m-n)/n^m

回答 (1)

用戶127502

2007-01-28 9:48 am

✔ 最佳答案

a) Inclusive-Exclusive Principle

P(A or B ) = P(A) + P(B) - P(A and B)
P(A|B|C) = P(A)+P(B)+P(C) - P(A&B)-P(B&C)-P(C&A) +P(A&B&C)
..
it can be generalized into n events.

Denote E_i as the event of i-th car empty
Required Probability
= 1 - P(E_1 or E_2 or .. or E_n)
and by Inclusive-Exclusive Principle,
P(E_1 or E_2 or .. or E_n)
= P(E_1)+P(E_2)+...+P(E_n) - [P(E_1 & E_2) + P(E_1&E_3) + ...] - ...
While P(E_i) is equal for all i, P(E_i & E_j) is equal for all i,j... etc
= nC1 x P(E_1) + (-1) nC2 x P(E_1&E2) + (-1)^2 ....
While P(E_1) = all m men go other n-1 car = [(n-1)/n]^m
, P(E_1&E_2) = all m men go other n-2 car = [(n-2)/n]^m

The result then followed.

b) It is equivalent to:
Choose r cars, and there will be at least one man in remain n-r cars by assigning m men.
There are nCr way to choose r cars
So, put (n-r) as n in part a.
then you can get the result.

Actually I dunno why u compute your answer.

參考: me

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