A bag contains 4 balls. Two balls are drawn at random and are found to be white. What is the probability that all balls are white. this problem comes under the topic Baye's theorem. I have no clue how to solve this. 
I found this answer from one book,but not sure whether its correct


 A: The notation is not standard.   When they write $\mathsf P(\frac E A)$ the usual convention is to write $\mathsf P(E \mid A)$.   But some authors do use it.  And the solution is indeed correct.
What we know:   The bag contains four balls.   Two balls were drawn.   They were white.   (Thus the bag must have contained at least 2 white balls.)
This give the conditional probabilities.  $$\begin{align}\mathsf P(E \mid A) & = \dfrac{^2C_2}{^4C_2}\\ \mathsf P(E\mid B) & = \dfrac{^3C_2}{^4C_2} \\ \mathsf P(E\mid C) & = \dfrac{^4C_2}{^4C_2}\end{align}$$
What we assume:   Lacking any prior knowledge we must give equal weighting to the possibilities that the bag contains 2, 3, or 4 white balls.   Thus we assign the prior probabilities as: $\mathsf P(A)=\mathsf P(B)=\mathsf P(C)=\frac 1 3$.   This is pretty much just guestimation; but we must start somewhere.
Then we calculate the posterior probabilities from there using Baye's Theorem:
$$\mathsf P(C\mid E) = \frac{\mathsf P(E\mid C)\cdot\mathsf P(C)}{\mathsf P(E\mid A)\cdot\mathsf P(A)+\mathsf P(E\mid B)\cdot\mathsf P(B)+\mathsf P(E\mid C)\cdot\mathsf P(C)}$$
Remark   The validity of posteriors is wholly dependent on the assignment of the priors.   In essence we are improving a guess based on evidence.
