at least 10? that means you can get 2 or less wrong.
Pr(2 wrong)=(4/5)^2*(1/5)^10
pr(1 wrong)=(4/5)(1/5)^11
Pr(0 wrong)=(1/5)^12
so now, Pr( at least 10 right)=1-pr(2wrong)-pr(1 wrong)-pr(zero wrong)
It is a multiple choice exam with 12 problems, each problem has 5 options. You need to answer 10 correctly to pass, what is the probability that by guessing you will answer at least 10 problems correctly?
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