Expected Value of Intersection of two Binomial Random Variables Ok the problem is as follows: (I am currently studying for my first actuary exam so this isn't a specific hw question! Just trying to figure it out!)
A and B will take the same 10-question exam. 
Each exam question will be answered correctly by A with probability $0.7$, independently of her results on other questions.
Each exam question will be answered correctly by B with prob. $0.4$, independently of results on other questions and A.
a. Find the expected number of questions that are answered correctly by both A and B.
Ok... I started this by first calculating 
$$\mathbb E[A] = \sum_{i=1}^{10} i{10 \choose i}(0.7)^i (1-0.7)^{10-i}$$ -so, treating this as a binomial random variable
I did this same thing for E[B]. I got the following results:
E[A] = 7
E[B] = 3.570018...
Now... to get the expected number of questions answered correctly by both A and B I assumed that this must be the intersection of the probability that A and B will answer the same questions correctly. Since they are independent I did
E[A*B] = ∑i^2*(10 choose i)(.7)^i*(.4)^i*(1-.7)^(10-i)*(1-.4)^(10-i) for i=1,2,...,10
This answer gives me approximately 2.15, when the correct solutions is 2.8 according to the text book
Other thoughts I have had are:
-multiply only by i, instead of i^2 since this would provide the intersection of p(A) * p(B) (This makes more sense to me in reasoning), however this value would be smaller than 2.15... and therefor cannot be the correct either!
Calculate E[A], then for each E[A=i], calculate E[B] i from 1 up to that value of i and then sum. i.e= If A gets 4 questions right, calculate the probability that B gets 1 of those questions right, 2 of those questions right, 3 of those questions right, all of the same questions right. I haven't tried this on paper yet because I'm not really sure how to set it up? 
A little guidance, or if someone could enlighten me to where I am going wrong would be great!
Thank you in advance!
 A: The probability that both A and B correctly answer a particular question is $0.7\times0.4=0.28$, since those events are independent.  So you have $n=10$ trials with probability $p=0.28$ of success on each trial.  The expected value of a $\mathrm{Bin}(n,p)$ random variable is $np$, so in this case that is $10\times0.28$.
You have $(1-0.7)(1-0.4)$ where you need $1-((0.7)(0.4))$.
You refer to the "intersection of two binomial random variables" and the "intersection of probabilities".  But one takes intersections of events, not of random variables or numbers.
A: Remember, each question is an independent and identically distributed Bernoulli trial; that is, from each examinee's perspective, the outcome is either "correct" or "incorrect."  So the probability that a given question on the test is answered correctly by both A and B is simply the product of the individual probabilities of each answering correctly; i.e., $(0.7)(0.4) = 0.28 = p$.  Then the probability distribution for the number of questions that both examinees got correct is Binomial with parameters $n = 10$ and $p = 0.28$.
