Conditional Probability Question (Urns and Balls)

Question

The question follows:

Urn one contains two red, one black balls, urn two contains one red, three black balls, and urn three contains one red, one black balls. A student chooses urn one or urn two at random, and selects one ball from the chosen urn at random and transfers it into urn three. Then he draws a ball from urn three. Given that the ball he draws is red, what is the probability that the transferred ball is red?

My attempt:

Let $T_r$ be the event of transferring a red ball to the 3rd urn and $F_r$ be the even of picking a red ball from the 3rd urn. We want to find $P(T_r|F_r)$.

We know that $P(T_r|F_r)=\frac{P(T_r\cap F_r)}{P(F_r)}$.

The numerator can be found by calculating the probability of transfering a red ball and from 2 red and 1 black ball picking the red one from the 3rd urn. The transfering part can be calculated by picking 1st or the 2nd urn with probability of 1/2 and then calculating the probability of picking a red ball and adding them together so that:

$P(T_r\cap F_r)=(1/2*2/3+1/2*1/3)(2/3)$

For $P(F_r)$, we can think it as $P(F_r)=P(T_r\cap F_r)+P(T_r^c\cap F_r)$ where $T_r^c$ is the event of transfering a blue ball.

So, $P(T_r^c\cap F_r)=(1/2*2/3+1/2*1/3)(1/3)$

Which gives us $P(T_r|F_r)=0.6666..$

I am trying to figure out probability with baby steps, is my answer correct? If no, can you explain it to me clearly? Thanks a lot in advance.

Answer 1

$ P(T_r|F_r)=\frac{P(T_r\cap F_r)}{P(F_r)}=$ $\frac{\frac{1}{2}\frac{2}{3}\frac{2}{3}+\frac{1}{2}\frac{1}{4}\frac{2}{3}}{(\frac{1}{2}\frac{2}{3}\frac{2}{3}+\frac{1}{2}\frac{1}{3}\frac{1}{3})+(\frac{1}{2}\frac{1}{4}\frac{2}{3}+\frac{1}{2}\frac{3}{4}\frac{1}{3})}=\frac{22}{35}$

$(\frac{1}{2}\frac{2}{3}\frac{2}{3})$transfering a red from urn$_1$ and getting a red from urn$_3$

$(\frac{1}{2}\frac{1}{3}\frac{1}{3})$transfering a blue from urn$_1$ and getting a red from urn$_3$

$(\frac{1}{2}\frac{1}{4}\frac{2}{3})$transfering a red from urn$_2$ and getting a red from urn$_3$

$(\frac{1}{2}\frac{3}{4}\frac{1}{3})$transfering a blue from urn$_2$ and getting a red from urn$_3$

Answer 2

$P(T_r|F_r)= \frac {P(F_r|T_r) P(T_r)}{P(F_r)}$ ...(i)

What are each of them?

$P(T_r|F_r)$ is the conditional probability that the transferred ball was red given we fetched red.

$P(F_r|T_r)$ is the conditional probability that we would fetch a red ball if the transferred ball was red.

$P(T_r)$ is the probability of transferring a red ball.

$P(F_r)$ is the probability of fetching a red ball.

Now if the transferred ball is red, we will have $2$ red balls and $1$ black ball in the third urn.

So, $P(F_r|T_r) = \frac{2}{3}$ ...(ii)

Now, as the probability of picking a red ball from urn $1$ is $2/3$ and from urn $2$ is $1/4$ and probability of picking one of these urns is $1/2$,

$P(T_r) = \frac{1}{2} (\frac {2}{3} + \frac {1}{4}) = \frac{11}{24}$ ...(iii)

Total number of balls in third urn after the transfer = $3$.

Expected number of red balls in urn $3$ after the transfer = $1 + \frac{11}{24} = \frac{35}{24}$

So $P(F_r) = \displaystyle \frac {\frac{35}{24}}{3} = \frac{35}{72}$ ...(iv)

Now substituting values from (ii), (iii), (iv) in (i)

$P(T_r|F_r)= \displaystyle \frac {\frac{2}{3} \times \frac{11}{24}}{\frac{35}{72}} = \frac{22}{35}$