William is taking the 25-question, multiple choice American Mathematics Competition. Each question has five answer choices. William guesses random answers for the last four questions. What is the probability that he will get at least two of these final four questions right?
Four questions are guessed randomly. Each question has 5 choices, one of which is correct.
So there are 625 permutations of answers and a 1/625 chance of getting every question right.
The easiest way to solve this is to use complementary counting. Find the number of cases where he gets 1 correct or none correct.
There is a 4/5 chance of getting each question WRONG. So the probability of getting them all wrong is 256/625.
The probability of getting ONE AND ONLY ONE correct is 1/5*4/5*4/5*4/5(the order doesn't matter here)=64/625.
Adding them up together, we get 320/625. But these are only the cases that we don't want, so we have to subtract from the total.
625-320=305. So the answer is 305/625=61/125.
I will leave you to check any possible errors. But you get the idea