Calculating Probability of Event After Y Attempts

[Drakkith]

TL;DR

Overall chance that A will occur given X chance per attempt and Y number of attempts.

Hey all. I've got next to no education in probability and I was wondering how to figure out the chance of some event occurring after Y number of attempts given X chance of happening per attempt.
For example, if event A has a 0.15% chance of occurring each attempt, and you make, say, 1000 attempts, what is the chance that A will occur?
Hope that makes sense.
Thanks in advance.

 

[jedishrfu]

Wouldn't that be

0.15% = 0.0015

1000 x 0.0015 = 1.5 times

 

[FactChecker]

Assume that all the attempts are independent. The probability of $A$ occurring at least once would be $1-ProbabilityThatANeverOccursIn1000 = 1-(1-0.0015)^{1000} = 0.77712$

 

[FactChecker]

Wouldn't that be

0.15% = 0.0015

1000 x 0.0015 = 1.5 times

Yes, that would be the expected number of occurrences of $A$, assuming that all the attempts are independent. The probability of at least one occurrence of $A$ is 0.77712

 

[PeroK]

Summary: Overall chance that A will occur given X chance per attempt and Y number of attempts.

Hey all. I've got next to no education in probability and I was wondering how to figure out the chance of some event occurring after Y number of attempts given X chance of happening per attempt.
For example, if event A has a 0.15% chance of occurring each attempt, and you make, say, 1000 attempts, what is the chance that A will occur?
Hope that makes sense.
Thanks in advance.

The best approach is to calculate the probability that A will not occur. If each trial is independent and $p$ is the probability that A occurs in anyone trial, then the probability that A does not occur in $n$ trials is $(1-p)^n$.

The probability that A will occur is then the complement of this, I.e $1-(1-p)^n$.

 

[Drakkith]

Thanks all!

 

[PeroK]

Note that the probability that A does occur includes all the cases where A occurs once, twice, three times up to all $n$ times. For that reason it's usually harder to calculate directly.