Something with a probability of 1% occurring 100 times I was thinking today that if something with a probability of occurring of 1% happened 100 times, then the probability of that something happening is 100%, I believe that according to the addition rule for probabilities the probabilities for each event should be added up to get the total probability thus 1/100 + 1/100 + 1/100 ... up to 100 = 100/100 = 1 = 100%.
Now, there's still the possibility that the event didn't occur any one of those 100 times when it could have, because each time is independent. If such is the case, then obviously the probability is not 100%.
I believe I'm wrong and that I'm doing something wrong. So I would very much appreciate any guidance as to how to go about calculating the probability of something that happens 100 times that has a chance of occurring of 1% every time.
For example, let's say there's a probability of 1% of dying from eating too much Cap'n Crunch, if I ate too much Cap'n Crunch 100 times, what is the probability that I will die?
 A: Think about it this way: The probability of not happening is .99, so each time, p = p x 0.99. P of never happend in 100 times is 0.99 ^ 100 = 0.366. Finally, the probability of a 1% 100 times happened at least once is 0.63.
A: In order to calculate the probability of at least one successful experiment out of $n$ experiments, you should calculate $1$ minus the probability of the complementary event (i.e., $1$ minus the probability of no successful experiment out of $n$ experiments).
The general formula is $1-(1-p)^{n}$, where $p$ is the probability of success in a single experiment.
In your question $p=0.01$ and $n=100$, hence the answer is $1-(1-0.01)^{100}\approx63.39\%$.
It is worth noting that in order for this method to be correct, the experiments must be independent of each other (i.e., the result of any experiment must not impact the result of any other experiment).
A: If two events A and B are mutually exclusive (i.e. it's possible for just A to occur, or for just B to occur, but never A and B together), then the probability of either A or B occurring is the sum of the individual probabilities - i.e. $P(A \lor B) = P(A) + P(B)$. That's the additivity of probabilities that you might be thinking of.
So that means that, for example, if you roll a 100-sided die, then the probability of any individual value is 1%, i.e. $P(1) = P(2) = P(3) = \ldots = P(100) = 0.01$. Because those events are exclusive (if the die roll is a 17, it can't also be 98). And the total of all of them, which is the probability of rolling 1 or 2 or 3 or ... or 100, is $P(1) + P(2) + \ldots + P(100) = 100 \times 0.01 = 1$.
However, for independent events (i.e. ones where the outcome of one doesn't affect the other), the probability of both occurring is the product of their individual probabilities. For example, the probability of rolling a 56 on my 100-sided die, then getting heads when I flip a fair coin, is $P(56 \land H) = P(56) \times P(H) = 0.01 \times 0.5 = 0.005$, i.e. 0.5%. Similarly, on two separate rolls of the die, the probability of getting 56 and then 21 is $0.01 \times 0.01 = 0.0001$.
So, if the probability of some event is 1%, and it has 100 chances to happen (for example, I roll my 100-sided die 100 times, and see if I ever roll 100), then we figure it as such:


*

*The probability of rolling 100 in one roll is 0.01, so the probability of not rolling 100 in one roll is 0.99.

*The probability of not rolling 100 in 100 rolls is the probability that it isn't 100 the first roll, and it isn't 100 the second roll, and it isn't 100 the third roll, ..., and is isn't 100 the hundredth roll. So the probability of that is $0.99 \times 0.99 \times \ldots \times 0.99 = 0.99^{100} \approx 0.366$.

*The probability of getting at least one roll of 100 is equal to one minus the probability of never getting a roll of 100 (because the two events are mutually exclusive and between them they describe all possible outcomes, so the sum of their probabilities must be 1). So its probability is $1-0.366 = 0.634$, or 63.4%.

