Answer to Question #127355 in Statistics and Probability for Emmason

Question

A man answers 10 maths problems one after the other. He answers the first problem correctly and the second incorrectly, for each of the remaining 8 problems the probability that he answers the problem correctly equals to the ratio of the number of problems that he has already answered correctly to the total number of problems that he has already answered. What is the probability that he answers exactly 5 out of 10 problems correctly?

Answer 1

The probability that he answers the problem 3 correctly equals

$p_3={1\over 2}$

The probability that he answers the problem 4 correctly equals

$p_4={1 \over 2}\cdot{2 \over 3}+{1 \over 2}\cdot {1 \over 3}={1 \over 2}$

The probability that he answers the problem 5 correctly equals

$p_5={1 \over 4}\cdot{3 \over 4}+{1 \over 4}\cdot {2 \over 4}+{1 \over 4}\cdot{2 \over 4}+{1 \over 4}\cdot {1 \over 4}={1 \over 2}$

Then the probability that he answers some problem correctly equals $\dfrac{1}{2}$ .

Let X= the number of questions correctly solved.

We have the binomial distribution: X$\sim$ Bin(n, p)

n=10, p=$\dfrac{1}{2}$

$P(X=5)=\binom{10}{5}({1 \over 2})^5(1-{1 \over 2})^{10-5}={63 \over 256}=0.24609375$

​The probability that he answers exactly 5 out 10 problems correctly is approximately 0.246.