# Probability question

#### ishihara

##### New Member
In a certain family, the four children take turns washing dishes. Out of a total of four breakages, three were caused by the youngest child, who thereafter was called clumsy. Was it justified to call the child clumsy, or could such an occurrence be reasonably attributed to chance?

Attempted solution: 12/35 is the probability the youngest child broke three dishes.

#### asterisk

##### New Member
There are 35 ways the dishes can be broken. [math]{4 + 4 - 1}\choose{4-1}[/math] = [math]{8}\choose{3}[/math] = 35

There are three ways the youngest child can break three dishes. Youngest child breaks 3, and the remaining dish can be broken by any of the other three children.

12/35 is the probability any of the four children breaks 3 dishes.

Also if you also consider there are four ways one child can break all four dishes, there are 16/35 = .4571 probability any child could break 3 or more dishes.

So the youngest child is not necessarily clumsy, it could be random.