Probability of at least 3 identical numbers from an array I have an array {1, 2, 3, 4, 5} and I extract 4 random numbers. I want to find out the probability that at least 3 numbers are identical.
I concluded that it's the probability of having exactly 3 identical numbers added to the probability of having exactly 4 identical numbers.
The probability of having exactly 3 numbers is:  $$\binom{4}{3}*p^{3}*(1-p)^{4-3}$$
Now, my issue is, who is p? As far as I understand, p is: $$\frac{1}{5}$$
Since it represents the probability of extracting any random number, but I'm not sure.
 A: Let's see.
Number of ways to extract exactly three the same: $5\cdot 4 \cdot 4 = 80$ (five choices for the triple, four choices for the number different from the triple, four ways to place the non-triple number in the order).
Number of ways to extract exactly four the same: $5$.
Number of ways to extract any four: $5^4 = 625$.
So it looks like $85/625 = 17/125$ is the probability.
A: You calculated the probability of getting at least three of a $particular$ number. However, we can get at least three of $any$ of the five numbers.
Let $X$ denote the number of identical numbers. First we compute $P(X=3)$.
We must get three of one number and one of another. First choose one of the five numbers to be obtained three times. Then denote $p=\frac{1}{5}$. Thus we have
$$\begin{align*}
P(X=3)
&= {{5 \choose 1} \cdot {4 \choose 3} \cdot\frac{1}{5}^3 \cdot\frac{4}{5}}\\\\
&=0.128
\end{align*}$$
Next compute $P(X=4)$. Choose one number of the five to be selected all four times. We have
$$\begin{align*}
P(X=4)
&= {5 \choose 1} \cdot {4 \choose 4} \cdot\frac{1}{5}^4\\\\
&=0.008
\end{align*}$$
Finally, 
$$P(X=3)+P(X=4)=0.128+0.008=0.136$$
