I got the answer as 7/16 , can anyone confirm it? 
Box A contains 5 red and 3 white marbles Box b contains 2 red and 6 white marbles if a marble is drawn from each box, what is the prob that they are of same colour.

My solution:
RR  or WW
= (5/8)(2/8)  + (3/8)(6/8)
= 10/64 + 18/64
= 28/64
= 7/16
My teacher's solution:
Let E1 be the event that marble is from Box A and is Red:
P(E1)=(1/2)(5/8)= 5/16
Let E2 be the event that marble is from Box B and is Red:
P(E2)=(1/2)(2/8)= 1/8
Let E3 be the event that marble is from Box B and is White:
P(E3)=(1/2)(3/8)= 3/16
Let E4 be the event that marble is from Box B and is White:
P(E4)=(1/2)(6/8)= 3/8
Required Probability= P(E1nE2)+P(E3nE4)= (5/16)(1/8)+(3/16)(3/8)= 7/64
can anyone confirm which one is correct?
 A: Your teacher is wrong (or at least has a different interpretation of the random experiment that I fail to read into the problem statement). They computed a different probability:

Randomly select a box and then randomly draw a marble from the selected box. Do the same a second time (independently). What is the probability that the the marbles have the same colour and the first marble was from the first box and the second marble was from the second box.

In particular, they seem to anticipate the possibility that the marbles might be chosen from the same box, whereas the problem statement explicitly says otherwise.
A: You are right and your teacher is wrong.
I think the easiest way to see that is to compare the arguments in the simple case when each box has just two balls, one of each color. Then the experiment is just like flipping two coins and asking whether the results match. That's $1/2$, which is what your reasoning leads to.
A: The probability that they are both white is obviously $\frac38\cdot\frac68=\frac{9}{32}$, which is already greater than $\frac{7}{64}$. So your teacher is clearly wrong.
A: 
Let $E_1$ be the event that marble is from Box A and is Red: $P(E_1)=(1/2)(5/8)= 5/16$

There is no need to multiply by $1/2$, the probability for selecting the box, as you are not randomly selecting boxes.
You are drawing a marble from each box, so that is certainly what will happen.
$E_1$ should be the event that the marble drawn from box A is red: $\mathsf P(E_1)=5/8$.
This is what you had calculated, and likewise  for the others.
So your calculation was correct.$$\dfrac{5}{8}\cdot\dfrac 28 + \dfrac 38\cdot\dfrac 68 = \dfrac{7}{16}$$
A: Firstly Box will be selected so that if he select box A then it will be 1/2.
Then as I had selected box A then I will drawn one ball the probability of getting red ball is 5/8 Similarly box B is selected then it will be 1/2.Then as I had selected box B then the probability of getting red ball is 2/8 hence the probability of getting same colour red ball is (1/2×5/8) +(1/2×2/8).
Secondly Box will be selected so that if he select box A then it will be 1/2.
Then as I had selected box A then I will drawn one ball the probability of getting white ball is 3/8 Similarly box B is selected then it will be 1/2.Then as I had selected box B then the probability of getting white ball is 6/8 hence the probability of getting same colour red ball is (1/2×3/8) +(1/2×6/8).
hence the final probability will be addition of it which will give us 7/64.
