Probability - two balls in the box: one we don't know its color and the other is red. What's the probability it's white? 
Bob has a black box (you can't see what's inside the box). A long time
ago Bob put one ball into the box but he doesn't remember what color
the ball was. With equal probability it can be a white ball or a red
ball. A. Bob takes a red ball and puts it in the same box.  Now
there are two balls in the box: one ball Red that Bob just put in and
a ball that was in the box earlier (Bob doesn't remember its color).
Now Bob draws Randomly one ball out of the box and it turned out to
be a red ball. Calculate the probability that the ball that has been
in the box for a long time is a white ball given the action taken by
Bob.

My attempt: There are two options, since we already know of them is red, $A_1= \{\text{White, Red}\}$ or $A_2= \{\text{Red, Red}\}$, so
$\Pr[A_1 \cup A_2] = 1/2$?
 A: You received two nice answers already, but let me add one that might be experienced as more intuitive.
If the ball has white color then the probability on drawing a red ball is $\frac12$.
If the ball has red color then the probability on drawing a red ball is $1$.
So comparing both situations you could say that the odds are $\frac12:1$.
We can also express that as $\frac13:\frac23$.
Here $\frac13+\frac23=1$ so that the two numbers can be interpreted as probabilities, and we can conclude that the first situation corresponds with probability $\frac13$.
A: Two possibilities:

*

*The first ball was white with probability $\frac12$.  The conditional probability of then drawing a red ball would be $\frac12$ so the joint probability would be $\frac12 \times \frac12=\frac14$.

*The first ball was red with probability $\frac12$.  The conditional probability of then drawing a red ball would be $\frac22=1$ so the joint probability would be $\frac12 \times 1=\frac12$.

This means the marginal probability of drawing a red ball was $\frac14 +\frac12  = \frac34$.
So by Bayes' theorem, given a red ball was drawn, the posterior probability that the first ball was white (and similarly that the remaining ball is white) is $\dfrac{\frac14}{\frac34}=\dfrac13$.
A: We are told that is it equally likely that Bob originally put a red ball in the box as it is that he put a white ball in the box.  However, the additional piece of information that a red ball has been selected makes it less likely that the box contains a white ball.
Let $E$ be the event that the box contains a red ball and a white ball; let $F$ be the event that the box contains two red balls; let $R$ be the event that a red ball is selected from the box.  Then we wish to calculate
\begin{align*}
\Pr(E \mid R) & = \frac{\Pr(E \cap R)}{\Pr(R)}\\
              & = \frac{\Pr(E)\Pr(R \mid E)}{\Pr(E)\Pr(R \mid E) + \Pr(F)\Pr(R \mid F)}\\
              & = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 1}\\
              & = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{2}}\\
              & = \frac{\frac{1}{4}}{\frac{3}{4}}\\
              & = \frac{1}{3}
\end{align*}
A: I look at it as a sequence of independent events probability.

*

*put in ball 1 as red (50%) or white (50%).

*put in ball 2 as red.

*pull out the first ball (50%), or pull out the second ball (50%).

This gives us 4 possible outcomes.

*

*(ball1=red, pulled=1) color of pulled ball = red

*(ball1=red, pulled=2) color of pulled ball = red

*(ball1=white, pulled=1) color of pulled ball = white

*(ball1=white, pulled=2) color of pulled ball = red

Only three of these are possible, due to observed ball being red.
Of these three cases only one has ball1=white.
So our chance is 1/3
