What is the probability of the drawn ball from U2 being white? This question is already posted here ,but i want to know flaw in my approach.
Question

Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls, and $U_2$ contains only $1$ white ball.

A fair coin is tossed:

*

*If head appears then $1$ ball is drawn at random from $U_1$ and put into $U_2$.

*If tail appears then $2$ balls are drawn at random from $U_1$ and put into $U_2$.

Now $1$ ball is drawn at random from $U_2$,What is the probability of the drawn ball from $U_2$ being white?
My Approach
There are $2$ cases.

*

*$\text{Case 1(Head Appears)}$
If Head Appears,then only $1$ ball will be drawn from $U_1$(white/red) to $U_2$
Current possible status of $U_2$
$WW,RW$
i.e $$\frac{1}{2}\times( \frac{2}{2}+\frac{1}{2} )$$

*

*$\text{Case 2(Tail Appears)}$
If Tail Appears,then  $2$ balls will be drawn from $U_1$(WW,WR,RR) to $U_2$
Current possible status of $U_2$
$WWW,WRW,RRW$
i.e $$\frac{1}{2}\times( \frac{1}{1}+\frac{2}{3} +\frac{1}{3})$$
Total Probability =$\text{Case 1+case 2}$
=$$\frac{1}{2}\times( \frac{2}{2}+\frac{1}{2} )+\frac{1}{2}\times( \frac{1}{1}+\frac{2}{3} +\frac{1}{3})$$
But my answer is coming $>1$,So it is absolutely wrong.But i am not getting where i am doing wrong.
Please help
 A: Following your approach it should be:
Case 1 (Head Appears): $U_2=WW$ with probability $\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$ and $U_2=RW$ with probability $\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}$.
Case 2 (Tail Appears): $U_2=WWW$ with probability $\frac{1}{2}\cdot\frac{3}{10}=\frac{3}{20}$, $U_2=RWW$ with probability $\frac{1}{2}\cdot\frac{6}{10}=\frac{6}{20}$  and $U_2=RRW$ with probability $\frac{1}{2}\cdot\frac{1}{10}=\frac{1}{20}$.
Hence the probability of the drawn ball from $U_2$ being white is:
$$\frac{3}{10}+\frac{1}{2}\cdot \frac{2}{10}+\frac{3}{20}+\frac{2}{3}\cdot \frac{6}{20}+\frac{1}{3}\cdot \frac{1}{20}=\frac{23}{30}.$$
A: Refer to the probability tree diagram:

Hence, the required probability is:
$$\frac12\cdot \frac35+\frac12\cdot\frac25\cdot\frac12+\frac12\cdot\frac35\cdot\frac24+\frac12\cdot\frac35\cdot\frac24\cdot\frac23+\frac12\cdot\frac35\cdot\frac24\cdot\frac23+\frac12\cdot\frac25\cdot\frac14\cdot\frac13=\\
\frac{18+6+9+6+6+1}{60}=\frac{46}{60}=\frac{23}{30}.$$
A: 
This diagram must help. The final Answer is 23/30 and not 23/50.
