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fugacity, 1 year ago to asklemmy in What are the odds of getting a passing grade by sheer guessing here?

P(passing) = 1- P(failure)
P(failure) = P(failure first try)*P(failure second try)
P(failure first try)=(3/4)^2
P(failure second try)=(gonna post in reply)

fugacity, 1 year ago to asklemmy in What are the odds of getting a passing grade by sheer guessing here?

his edit does not assume that; it's the cleanest way of doing the problem

fugacity, 1 year ago (edited 1 year ago) to asklemmy in What are the odds of getting a passing grade by sheer guessing here?

P(failure second try)=(2/3)^2 since you can eliminate one choice but 2 others are still wrong.

To total:
P(failure)=(3/4)^2*(2/3)^2=1/4
1-1/4=0.75

So the probability of passing is 0.75

Edit:
Remark: this problem is elegant if you attempt to calculate the passing as the complement of failure rather than enumerate all successes. Shouldn't take more than 3 minutes with a clear head if you know the correct approach. If this was an college level intro probability exam question, it should be done the fast way since it's meant to eat up your time otherwise.