$A$and $B$will take the same $10$-question examination. Each question will be answered correctly by $A$with probability$.7$, independently of her results on other questions. Each question will be answered correctly by B with probability $.4$, independently both of her results on the other questions and on the performance of $A.$

(a) Find the expected number of questions that are answered correctly by both A and B.

(b) Find the variance of the number of questions that are answered correctly by either A or B

The problem deals with the probability of two persons $A$and $B$answering correctly on a $10$question exam.

Probability of ' $A$ ' answering one question correctly,$P\left(A\right)=0.7$

The probability of success, that is, both $A$and $B$answering a question correctly is given by $p(A\cap B)=p\left(A\right)\times \mathrm{p}\left(B\right)$

$=0.7\times 0.4$

$=0.28.$

Let $X$denotes the number of questions that are answered correctly by both $A$and $B$on a $10$question exam.

Here,$X~Binom(n=10,p=0.28)$

The expected value is found as follows.

$E\left(X\right)=np$

$=10\times 0.28$

We get,

$=2.8.$

The expected number of correctly answered questions by both $A$and$B$on a $10$question exam is $2.8.$

The problem deals with the probability of two persons $A$and $B$answering correctly on a $10$question exam.

Probability of ' $B$' answering one question correctly,$P\left(B\right)=0.4.$

The probability that either $A$or $B$answers a question correctly is given below.

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=0.7+0.4-0.28$

We get,

$=0.82$.

Let$X$ denotes the number of questions that are answered correctly by either $A$or $B$on a$10$question exam.

Here, $X~Binom(n=10,p=0.82)$

Then, the variance of the number of questions correctly answered by either $A$or $B$is given by,

$Var\left(X\right)=np(1-p)$

$=10\times 0.82\times (1-0.82)$

$=1.476$

The variance for the number of correctly answered questions by either $A$or $B$is $1.476.$