SOLUTION: What is the probability of getting at least one correct answer on a 20-question test with 4 possible answers per question?
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Question 278989: What is the probability of getting at least one correct answer on a 20-question test with 4 possible answers per question?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The hard way to do this is to compute the probability of getting exactly 1 right, )
, then the probability of getting exactly 2 right, )
, then 3, then 4, and so on up to 20, and finally adding all of these probabilities together. All of that being a big pain in the sit-down because each of the calculations involves the following formula:
\ =\ \left(n\cr k\right)p^k\left(1\,-\,p\right)^{n-k})
Where 
is the number of trials, 
is the number of successes, 
is the probability of success on an individual trial, )
is the probability of failure on an individual trial, and )
is the number of ways to choose things from a collection of things and is equal to !)
.
All of which means that you have 19 complex calculations plus a 19-term addition problem. Ah, but there truly is a better way.
The opposite case of getting at least one right is getting none right. So the probability of getting at least one right plus the probability of getting none right is equal to 1. So, just calculate the probability of getting none of them right and subtract that from 1.
So:
\ =\ 1\ -\ P_{20}\left(0\right)\ =\ 1\ -\ \left(\frac{20!}{0!(20\,-\,0)!}\right)\left(\frac{1}{4}\right)^0\left(1\,-\,\frac{1}{4}\right)^{20-0}\ =\ 1\ -\ \left(\frac{3}{4}\right)^{20})
Hint: Remember 
and 
Also: 0.75 "x^y" 20 = on your Windows built-in calculator in Scientific mode will give you ^{20})
The rest of the arithmetic is yours to do.
John
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