Consider the set $S$ of partitions of four differently colored balls into as many as four groups.

What is the probability that in a partition chosen randomly from $S$ , the yellow ball is grouped alone?

$\dfrac{1}{5}$
$\dfrac{1}{2}$
$\dfrac{1}{6}$
$\dfrac{1}{4}$
$\dfrac{1}{3}$

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If we designate $R, B, G, Y$ as distinct elements for red, blue, green, & yellow respectively. If we desire a distinct set of yellow only, we will need to find out the partition sets of the other $3$ elements, and there are $5$ ways to arrange them:

{{R, B, G}}, {{R}, {B,G}}, {{B}, {R,G}}, {{G}, {B,R}}, {{R}, {B}, {G}}

From these $5$ arrangements, the yellow ball can be placed separately from the $3$ balls.

This partition number is also called Bell Number , and we can write the notation as:

$B_{3} = 5$ .

By adding $Y$ to join any of the $10$ sets above, we will obtain $10$ more arrangements and can calculate all possible outcomes for $4$ elements: $B_{4} = 15$

Therefore, the probability of only one yellow in one of the compartments = $\dfrac{B_{3}}{B_{4}} = \dfrac{5}{15} = \dfrac{1}{3}$ .