Wondering if it is merely Bayes Theorem? An urn contain two balls. It is known that the urn was filled by tossing a fair coin twice and putting a white ball for each head, and putting a black ball for each tail. A ball is drawn from the urn and is found to be white. What is the probability that the other ball in the urn is also white.
My Attempt:- Recently I encountered a similar problem where I applied Bayes Theorem. This question was actually asked in an interview. I was asked to write my answers in board. I tried in this way:-
Let $E_1, E_2,E_3$ be the events that the urn contains (1)two white balls, (2)two black balls and (3)a black and a white ball respectively. All three are mutually exclusive.
Now, 
$$P(E_1)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$
$$P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$
$$P(E_3)=2\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{2}$$
(I may get white in the first throw and black in the second. Or, black in the first and white in second.)
Let $A$ be the event that a ball is chosen and it is found to be white
$$P(A/E_1)=1$$
$$P(A/E_2)=0$$
$$P(A/E_3)=\frac{1}{2}$$
Now, we need $P(E_1/A)$.
By Bayes theorem,
$$P(E_1/A)=\frac{\frac{1}{4}\times 1}{\frac{1}{4}\times 1+\frac{1}{4}\times 0+\frac{1}{2}\times \frac{1}{2}}=\frac{1}{2}$$
Am I correct ?
 A: Yes, well, the working is correct, but unecessary.   You have revealed that you understand the math, but hadn't thought about the underlying problem.

"The probability is $1/2$.   Since the results of each coin toss -and therefore the colour of each ball- is independent of that of the other, then knowing the colour of that particular ball -so long as it was drawn from the urn without bias- tells me nothing of the colour of the unrevealed ball; other than that is was selected by the result of an independent fair coin toss."

A: Yes, your reasoning is correct. The denominator in your final calculation wasn't needed however. The first ball selected will be white with probability $\frac{1}{2}$.
$$P(E_1 \mid A)=\frac{P(A \mid E_1)\cdot P(E_1)}{P(A)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$$
An R simulation gives
x = replicate(10^6, sample(0:1,2,repl=T))
x=as.data.frame(x,row.names=c("first","second"))
x=t(x)
x=as.data.frame(x)
attach(x)
x <- x[ which(first==1),]
mean(second==1)

[1] 0.500002

