# variance and expected value

I have a problem with an exercise about the expected value and the variance.

John and Henry take an examination composed of 10 questions. John answers correctly every question with a probability 0.7, independently from the other questions. Henry answers correctly every question with a probability 0.4, independently from the other questions and independently from the answers given by John.

Calculate:

(a) The expected value of the number of questions correctly answered by both students.

(b) The variance of the number of questions that only one of the students answered correctly.

If we let $X$ equal the number of correct answers of John, then $X$ is a binomial random variable with parameters $n=10$, $p=0.7$. The expected value of the correct answers is $7$ while the variance of $X$ is $np(1-p) = 10\times 0.7\times 0.3= 2.1$ for John.

The same way, if we let $Y$ equal the number of correct answers of Henry, then $Y$ is a binomial random variable with parameters $n=10$, $p=0,4$. The expected value of the correct answers is $4$ while the variance of $X$ is $np(1-p) = 10\times 0.4\times 0.6= 2.4$ for Henry.

The solutions are then $2.8$ and $1.476$.

Let $P(J_1)$ and $P(H_1)$ be the probability that John and Henry answer a question CORRECTLY, and let $P(J_0)$ and $P(H_0)$ be the probability that John and Henry answer a question INCORRECTLY.
The probability that a given question is correctly answered by both students is $P = P(J_1) \times P(H_1) = 0.7 \times 0.4 = 0.28$. Intuitively, this means that John and Henry BOTH have to answer the question correctly. You now have a binomial with $n = 10$ and $p = 0.28$.
Now we want to consider the probability that a given question is correctly answered by only ONE of the two students. This can occur in one of two circumstances: i) John answers the question correctly and Henry answers the question incorrectly, or ii) Henry answers the question correctly and John answers the question incorrectly. Therefore, the probability of this event is: $P = P(J_1)P(H_0) + P(J_0)P(H_1) = 0.7 \times 0.6 + 0.3 \times 0.4 = 0.54.$ You now have a binomial with $n = 10$ and $p = 0.54$.
• @Anne Note that the solution that you gave for part (b) refers to the case where at least one of the students gets the answer correctly (i.e. $p$ in that case is equal to 0.82, which means that $\sigma^2 = n p (1-p) = 10 \times 0.82 \times 0.18 = 1.476)$. Are you sure that you formulated the question correctly? – Michael R Dec 17 '16 at 20:51
• Thank you for your answer. Actually I have some translation problems with problems from a book (a bad translation from an American book). I've tried with $P=P(J_1)P(H_0)+P(J_0)P(H_1)+P(J_1)P(H_1)=0,82$ and the solution is good – Anne Dec 17 '16 at 21:15