Finding probability of ball being white using Bayes theorem? From an urn containing 3 white and 5 black balls, 4 balls are transferred into an empty urn. From this urn 2 balls are taken and they both happen to be white. They are then replaced. What is the probability that the third ball taken from the same urn will be white?
The answer seems to be 7/12, but my answer is coming to be 15/28.
Here is my approach so far:
Since 2 balls were found to be white and replaced already, at least 2 balls out of 4 are white, so probability will be >50%. 
The two possibilities are: 2 white + 2 black, or 3 white + 1 black.
So the probability of another ball being white will be: 3/4x1/7 + 2/4x6/7 = 15/28
 A: Let $X$ denote the number of white balls initially transferred into the urn from which you are drawing balls.  Then, before you have drawn any balls, $X$ has a hypergeometric distribution:  $$\Pr[X = x] = \frac{\binom{3}{x}\binom{5}{4-x}}{\binom{8}{4}}, \quad x \in \{0, 1, 2, 3\}.$$  After you draw two balls from this urn and note they are white, then it becomes clear that $X \in \{2, 3\}$:  We are assured at least two of the four balls is white, and as there were only three white balls from which the second urn is populated, it is impossible to have more than three in the second urn.  This much you have noted.  However, what is the posterior probability of $X = 2 \mid Y = 2$?  Well, Bayes theorem gives us $$\Pr[X = 2 \mid Y = 2] = \frac{\Pr[Y = 2 \mid X = 2]\Pr[X = 2]}{\Pr[Y = 2]}.$$  The numerator is simple:  If there were only two white balls in the second urn, the chance that you draw both of them is $1/6$, since there is only one way to pick both white balls out of the $\binom{4}{2}$ ways to take two balls without replacement.  So the numerator is simply $$\Pr[Y = 2 \mid X = 2]\Pr[X = 2] = \frac{1}{6} \cdot \frac{\binom{3}{2}\binom{5}{2}}{\binom{8}{4}} = \frac{1}{14}.$$  The denominator is trickier, since we must calculate $\Pr[Y = 2 \mid X = 3]$.  In such a case, we can reason that the only way we can fail to obtain $2$ white balls is if one of the balls we pick is the black one, thus $$\Pr[Y = 2 \mid X = 3] = 1 - \frac{3}{6} = \frac{1}{2}.$$  Thus $$\Pr[Y = 2] = \Pr[Y = 2 \mid X = 2]\Pr[X = 2] + \Pr[Y = 2 \mid X = 3]\Pr[X = 3] \\ = \frac{1}{14} + \frac{1}{2}\cdot\frac{\binom{3}{3}\binom{5}{1}}{\binom{8}{4}} = \frac{3}{28}.$$  Therefore, $$\Pr[X = 2 \mid Y = 2] = \frac{1/14}{3/28} = \frac{2}{3},$$ and $$\Pr[X = 3 \mid Y =2] = \frac{1}{3}.$$
We then easily compute that the probability of obtaining a white ball from a single draw from this urn is $$\frac{2}{3}\cdot\frac{2}{4} + \frac{1}{3}\cdot\frac{3}{4} = \frac{7}{12}.$$
A: Four balls are transfered from an urn holding 3 white and 5 black balls.   Two balls are drawn from the new urn and are white and are then replaced; which I presume means they were first drawn without replacement.
Let random variable $X$ count the number of white balls transfered to the new urn, $E$ the event that two white balls are drawn from the new urn, and $F$ the event that a white ball will be drawn after the evidence is replaced.   $E$ and $F$ are conditionally independent given a value for $X$. 
As you observed, there are two plausible values of $X$ given the evidence that at least two white balls are drawn.   Yet you neglect to evaluate the probabilities, and for this Bates' Theorem is useful.
$\mathsf P(X{=}2)=\left.\binom 32\binom 52\middle/\binom 84\right.$ is the probability for selecting two from three white balls and two from five black balls when selecting foutfrom eight balls in the first urn.
$\mathsf P(E\mid X{=}2)= \left.\binom 22\binom 20\middle/\binom 42\right.$ is the probability for selecting two from the white balls and none from the black balls when selecting two from four balls given two of them are white. 
$\mathsf P(F\mid X{=}2)= \left.\binom 21\binom 20\middle/\binom 41\right.$ is the probability for selecting one from the white balls and none from the black balls when selecting two from four balls given two of them are white.
And so forth.   Then by the Law of Total Probability, the conditional independence, and Bayes' Theorem:
$$\begin{align}\mathsf P(F\mid E)&=\mathsf P(F\mid X{=}2)\,\mathsf P(X{=}2\mid E)+\mathsf P(F\mid X{=}3)\,\mathsf P(X{=}3\mid E)\\[1ex]&=\dfrac{\mathsf P(F\mid X{=}2)\,\mathsf P(E\mid X{=}2)\,\mathsf P(X{=}2)+\mathsf P(F\mid X{=}3)\,\mathsf P(E\mid X{=}3)\,\mathsf P(X{=}3)}{\mathsf P(E\mid X{=}2)\,\mathsf P(X{=}2)+\mathsf P(E\mid X{=}3)\,\mathsf P(X{=}3)}\\[1ex]&~~\vdots\\[1ex]&=\dfrac{7}{12}\end{align}$$
