Probability of an event occurring at least once over n trials

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The probability of Joe contracting an STD at least once over four sexual encounters, given a 1 in 4 chance per encounter, can be calculated using complementary probability. Instead of adding probabilities directly, the probability of not contracting an STD in a single encounter is 3 in 4. Therefore, the probability of not contracting an STD in all four encounters is (3/4)^4. Subtracting this value from 1 gives the probability of contracting an STD at least once. This approach clarifies the misunderstanding of simply summing probabilities.
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Homework Statement


There is a 1 in 4 chance that on any given sexual encounter, Joe will contract an STD. If Joe has sex 4 times, what are the chances of him having an STD?



Homework Equations



Something with combinations maybe?

The Attempt at a Solution



I thought you would just add up the probability of the event happening for each trial, but that would give me 100% which isn't right.
 
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helpmepleasezz said:

Homework Statement


There is a 1 in 4 chance that on any given sexual encounter, Joe will contract an STD. If Joe has sex 4 times, what are the chances of him having an STD?



Homework Equations



Something with combinations maybe?

The Attempt at a Solution



I thought you would just add up the probability of the event happening for each trial, but that would give me 100% which isn't right.

Something simpler. What is the complementary event to "getting STD from at least one encounter"?
 

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