# Calculating Probability of Event After Y Attempts

• B
• Drakkith

#### Drakkith

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TL;DR 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.

• hutchphd
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 ##

• Drakkith
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

• hutchphd, jedishrfu and Drakkith
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.
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##.

• hutchphd and Drakkith
Thanks all!

• hutchphd
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.

• hutchphd