Zachary Agatstein 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.
I agree.
> But the restriction for the 4th apple to only be limited to baskets 1
> thru 4 would obviously change that probability. How?
You can ask: If you have a valid solution to the simplified problem (ie
"there is no more than
one apple in every basket"), what is the probability that apple No 4 was
thrown into one of the baskets 1-4?
It's 4/8 = 0.5, so
0.5*0.41015625 = 0.205078125
is the probability you've searched.
Tobias Arens
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