what is the probability that at least 1 number is prime in the following experiment? Imagine that you have the first 100 numbers and you pick up one randomly, then you write down which one it is and put the number again in the set of the first 100 numbers. You repeat this experiment 30 times. What is the probability that at least one of these numbers is a prime number?
My approach to this question is the following:
There are 25 primes numbers in the first 100 numbers, therefore the probability of picking up a prime is $P(p)=\frac{25}{100}=0.25$. So if you do this 30 times you have to sum up all the probabilities $$P(p_1)+P(p_2)+\cdots+P(p_{30})=\frac{25}{100}+\frac{25}{100}+\cdots+\frac{25}{100}=30\cdot\frac{25}{100}$$ but this approach is obviously wrong because this probability is grearter than 1. What am I missing?
 A: The probability that the first one is prime is independent from the rest. So, the probability that the first choice is prime is $\dfrac{25}{100}$, but that is true whether the second one is prime or not. So that probability is counting when the second choice is prime and when it is composite. Similarly, the probability that the second choice is prime is $\dfrac{25}{100}$, but that is independent of whether the first was prime or not. So, it is already counting when the first is prime and when it is not. So, when you add the two probabilities, you are including the possibility that both are prime multiple times. In combinatorics, this would be called "overcounting". I am not sure what the term is in probability theory.
One possible solution is to subtract the overcounting. This can be accomplished via Inclusion/Exclusion. The formula for this would be very messy.
However, there is an easier approach as @lulu mentioned. Consider the opposite of at least one number selected is prime. The opposite would be that every number selected is composite. By the law of total probability, we have:
$$P(\text{select at least one prime}) + P(\text{all selections are composite}) = 1$$
Solving for $P(\text{select at least one prime})$ gives:
$$P(\text{select at least one prime}) = 1 - P(\text{all selections are composite})\\ = 1 - \left(\dfrac{75}{100}\right)^{30}$$

This type of repeated random event is known as a Bernoulli trial. The probability that you choose exactly $k$ primes among the thirty trials is given by:
$$\dbinom{30}{k}\left(\dfrac{25}{100}\right)^k\left(1-\dfrac{25}{100}\right)^{30-k}$$
So, another way to calculate the probability of at least one prime is the probability of exactly one prime, plus the probability of exactly two primes, plus the probability of exactly three primes, etc.:
$$\sum_{k=1}^{30}\dbinom{30}{k}\left(\dfrac{25}{100}\right)^k\left(1-\dfrac{25}{100}\right)^{30-k}$$
To see that this yields the correct answer again, consider the law of total probability. The only outcome we are not including is the probability of choosing zero primes. So, we expect:
$$\sum_{k=0}^{30}\dbinom{30}{k}\left(\dfrac{25}{100}\right)^k\left(1-\dfrac{25}{100}\right)^{30-k} = 1$$
I plugged it into Wolframalpha, replacing your probability $\dfrac{25}{100}$ with a generic probability $p$. And instead of $30$, I used $n$ trials. Feel free to replace those values back, and you will get the same result:
Result from Wolframalpha
