# Probability to choose all the numbers

The problem

Let's consider the numbers $1$, $2$ and $3$. Five people have to choose secretly one of those numbers.

I want to find the probability that every number has been been chosen.

What I did

Let $A_k$ be the even "the number $k$ is chosen.

So I think we have

$$\mathbb P({}^cA_k)=\left(\frac 23\right)^5.$$

The event I am interested in should be

$$A_1\cap A_2\cap A_3,$$

and

$$\mathbb P(A_1\cap A_2\cap A_3)=1-\mathbb P({}^c(A_1\cap A_2\cap A_3))$$

$$\mathbb P(A_1\cap A_2\cap A_3)=1-\mathbb P({}^cA_1\cup {}^cA_2\cup {}^cA_3))$$

$$\mathbb P(A_1\cap A_2\cap A_3)=1-\mathbb P({}^c A_1)-\mathbb P({}^c A_2)-\mathbb P( {}^c A_3)+\mathbb P({}^c A_1\cap{}^c A_2)+\mathbb P({}^c A_1\cap{}^c A_3)+\mathbb P({}^c A_3\cap{}^c A_2)-\mathbb P({}^c A_1\cap{}^c A_2\cap{}^c A_3)$$

$$\mathbb P(A_1\cap A_2\cap A_3)=1-3\left(\frac 23\right)^5+3\left(\frac 23\right)^{10}-\left(\frac 23\right)^{15}$$

$$\mathbb P(A_1\cap A_2\cap A_3)=\frac{9393931}{14348907}\approx 0.654679.$$

Though the result should be $0.61728395$ according to the correction.

What did I do wrong ?

You are correct that the probability a particular number is not chosen is $\left(\dfrac23\right)^5$
So instead you should use a counting argument: there are $3^5$ ways of choosing the numbers, $3\times 1^5$ of which have a single number chosen and $3\times 2^5 -2\times 3\times 1^5$ have exactly two numbers chosen. So that makes the probability of exactly three numbers being chosen $$\dfrac{3^5 - 3\times 1^5 -( 3\times 2^5 -2\times 3\times 1^5)}{3^5}$$
If 1 is not chosen in this setup, you have triplets like ($2,2,2), (2,2,3)...(3,3,3)$. What s the probability to get this outcome? Now consider tge fact that you have 3 such n7mbers.