Photo:
THE MIRROR
There are five buckets on the floor.
Five people can each throw a ball into one of the buckets.
The selection of the bucket is random - it does not depend on whether a bucket is empty or already has at least one ball in it.
What is the probability that after the five balls have been distributed, exactly one of the five buckets will be empty?
The probability is
1200/3125 = 0.38...
A total of 55 = 3125 different distributions are possible because each of the five balls can land in one of the five buckets.
If there is exactly one ball in each bucket, there are 5*4*3*2*1 possibilities.
Why?
Let's imagine that the balls are thrown in one after the other.
The first person has five buckets to choose from.
Person 2 only four, because one bucket is already occupied.
Person 3 has three options, person 2 has two, and the last person has only one.
This results in the product 5*4*3*2*1, which mathematicians call the factorial of 5.
5!
is the mathematical notation.
However, there must be two balls in one bucket and one each in three others for exactly one bucket to be empty.
Four people throw a ball into a previously empty bucket.
As in the example above, there are 5*4*3*2 different distributions for the four people.
A person catches a bucket in which there is already a ball.
This cannot be the person who throws the first ball into a bucket, but each of the four following people.
Person 2 has exactly one option to put the ball in a bucket that already has a ball in it.
Person 3 has two options, person 4 has three, and person 5 has four.
This results in the following total number of distributions:
Z = 5*4*3*2 * (1+2+3+4)
Z = 120*10
Z = 1200
The probability is therefore 1200/3125.
I discovered this puzzle on the Mathigon portal.
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