The probability that he answers the problem 3 correctly equals
p3=21 The probability that he answers the problem 4 correctly equals
p4=21⋅32+21⋅31=21
The probability that he answers the problem 5 correctly equals
p5=41⋅43+41⋅42+41⋅42+41⋅41=21 ...
Then the probability that he answers some problem correctly equals 21.
Let X= the number of questions correctly solved.
We have the binomial distribution: X∼Bin(n,p)
n=10,p=21
P(X=5)=(510)(21)5(1−21)10−5=25663==0.24609375The probability that he answers exactly 5 out 10 problems correctly is approximately 0.246.
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