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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

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marked as duplicate by MJD, voldemort, Tunk-Fey, amWhy, Mark Fantini Aug 27 '14 at 0:23

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    $\begingroup$ Zero. The other two balls can be purple, green, orange, black.. :) $\endgroup$ – Jack D'Aurizio Aug 26 '14 at 22:40
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    $\begingroup$ We have to choose a "prior," and the wording of the problem does not give any guidance. If we decide that the $4$ balls that were placed in the bag were chosen, independently, with white and red equally likely, then one can produce an answer. But there is no good reason to use this as the prior. $\endgroup$ – André Nicolas Aug 26 '14 at 22:40
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    $\begingroup$ I would think your exercise would specify some a priori distribution of the balls in the urn, which you would then update using the outcomes of the first two draws. $\endgroup$ – hardmath Aug 26 '14 at 22:40
  • $\begingroup$ I'm just echoing what has been said but still, the problem as it is stated just doesn't give enough information to calculate a probability at all. $\endgroup$ – bob.sacamento Aug 26 '14 at 22:41
  • $\begingroup$ At least we're given that the bag contained 4 balls! What if it had been "Two white balls are drawn from a bag. What is the probability that the bag contains two further white balls?" $\endgroup$ – MJD Aug 26 '14 at 23:29
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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.

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