In article <[EMAIL PROTECTED]>,
Zachary Agatstein <[EMAIL PROTECTED]> wrote:
>Can you help me solve this problem:
>There are 8 baskets and 4 apples. Thrown at random, 3 of the 4 apples
>can go to any basket. The 4th apple, however, can only be thrown into
>baskets 1 through 4. What is the probability that there is no more than
>one apple in every basket?
>Now, I can easily solve this problem if the 4th apple could also go to
>any of the 8 baskets. The probability referred to above can be computed
>as follows:
>P = factorial(8)/((8 to the power of 4)*factorial(8-4)) = 0.41015625.
>But the restriction for the 4th apple to only be limited to baskets 1
>thru 4 would obviously change that probability. How?
A simple argument, without calculation, shows that the probability
is exactly the same. For whatever basket the fourth apple is in,
the probability is exactly the same that none of the other three
apples go to that basket, or to the same basket as any other.
Now if two apples are not equally likely to go to each basket,
and are not independent, this argument fails.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
[EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558
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