Htykuut

I'm thinking about the following:

I have 3 types of things, 3 identical pieces of each. Each of 5 people randomly choose 1 item. What is the probability that there still is at least 1 item left from each type?

I was thinking about combination with repetition, which however doesnt consider who gets what.

Thank you in advance.

Let's call your items $a_{11}$ through $a_{33}$, with $a_{11},a_{12},a_{13}$ being of type $t_1$, same for $t_2$ and $t_3$.

There are ${9choose 4}$ equiprobable final configurations after 5 items have been selected. Let's compute the number of configurations where at least one item of each kind is left.

There are 3 cases we want to look at:

• 
  $C_1$ : 2 of $t_1$, 1 of $t_2$, 1 of $t_3$
  


• 
  $C_2$ : 1 of $t_1$, 2 of $t_2$, 1 of $t_3$
  


• 
  $C_3$ : 1 of $t_1$, 1 of $t_2$, 2 of $t_3$
  

They're mutually exclusive and symmetric so we can just compute the first case and the end result is 3 times that.

To get a case $C_1$, we remove one of three from $t_1$, we keep one of three from $t_2$, we keep one of three from $t_3$, so the number of cases $C_1$ is $N(C_1) = 3 * 3 * 3 = 3^3$.

The number of admissible configurations is then $3N(C_1) = 3^4$, which means the probability of getting an admissible configuration is :

$$frac{3^4}{{9choose 4}} = frac{9}{14}$$

Number the types with $i=1,2,3$.

Let $E$ denote the event that still $1$ item of each kind is left.

Let $A_i$ denote the event that no items are left from the $i$-th type.

Then: $$E^{complement}=A_1cup A_2cup A_3$$

Moreover the events $A_1,A_2,A_3$ are mutually exclusive and equiprobable so that $$P(E)=1-P(E^{complement})=1-P(A_1)-P(A_2)-P(A_3)=1-3P(A_1)$$

Here: $$P(A_1)=frac{binom33binom62}{binom95}=frac5{42}$$ corresponding with a selection of $5$ items out of $9$ where exactly $3$ items of type $1$ are selected.

So we end up with: $$P(E)=1-3cdotfrac5{42}=frac{27}{42}=frac9{14}$$