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I have a set of 9 different elements. How do I calculate the chance that when I pick one element out of the set 20 times, I have the same one 5 times or more?

I thought about using the following formula: if the probability of success on any trial is $p$, then the probability of exactly $k$ successes in $n$ trials is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ (Source)

Then we have $p$ equal to $1/9$, $n$ equal to $20$ and $k$ equal to $5$ till $20$ and add up the values. We multiply this by $\binom{20}{4}$ so we take into account that there are $\binom{20}{4}$ ways to pick 4 elements out of 20.

Is this correct?

user2609980
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Not quite, I'm not sure where you got $\binom{20}{4}$ from. The probability that you get exactly $5$ successes out of $20$ is $$\binom{20}{5}\left(\frac{1}{9}\right)^5\left(\frac{8}{9}\right)^{15}.$$

However, you are interested in $5$ or more, and since the events are mutually exclusive, then the probability of interest is the sum of the probabilities of getting 5, or, 6, or 7,... or 20 successes; $$\sum_{k=5}^{20}\binom{20}{k}\left(\frac{1}{9}\right)^k\left(\frac{8}{9}\right)^{20-k}.$$

This is equivalent to $$1-P(4\text{ or fewer successes}) = 1-\sum_{k=0}^4\binom{20}{k}\left(\frac{1}{9}\right)^k\left(\frac{8}{9}\right)^{20-k}.$$

Em.
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