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 in four sexual encounters, given a 1 in 4 chance per encounter, is calculated by finding the probability of not contracting an STD in all four encounters, which is 1 - (0.75^4). The resulting probability indicates a significant risk of contracting an STD at least once. To determine the probability of Joe contracting an STD exactly twice in those four encounters, a combination formula would be necessary. This involves using the binomial probability formula, which accounts for the number of successes in a given number of trials. The discussion highlights the application of probability concepts in real-life scenarios.
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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?
 
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Hint: what is the probability of not contracting after 4 times?
 
1-(.75*.75*.75*.75) = my answer?
 
Looks good.
 
So how would you do it for the probability of Joe contracting an STD twice out of those 4 occasions?

I think you would have to do a combination for this?
 
As this is misplaced homework-type question, I am locking it. This thread was also posted in the Precalc section under Homework & Coursework.
 

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