There are $100$ yellow balls and $300$ blue balls in urn I, and there are $300$ yellow balls and $100$ blue balls in urn II. You take half of the balls from I and transfer them to II. Then you chose one ball from II, what is the probability that this ball is yellow? Let $\mbox{Urn}_i$ (Yellow, blue) denote the number of Y and P balls in urn $i$.
So the first thing I did was use smaller numbers to get a better intuition.
Time $0$: $I(1,3) \quad II(3,1)$
Transfer Case $1$ : $(1,1)$ from $I \to II \quad$ then $I(0,2) \quad II(4,2)$
Transfer Case $2$ : $(0,2)$ from $I \to II \quad$ then $I(1,1) \quad II(3,3)$
If $Y$ denotes the event that yellow is chosen from $II$ after the transfer, then
$Pr(Y)=Pr(Y|(1,1) \mathrm{\ transfer})+Pr(Y|(0,2) \mathrm{\ transfer})$
$Pr(Y|(1,1) \mbox{ transfer})=\frac 2{(4+2)}= \frac 13$ from the final numbers in $II$ after the transfer Case 1 $Pr(Y|(0,2) \mathrm{\ transfer})=\frac 3{3+3}= \frac 12$
So $Pr(Y)=\frac 56$ using the smaller scaled problem. Now when I scale this to a larger multiples...I don't understand the pattern to get a better intuition.
I found this problem and I thought it was very similar, but I'm struggling with how to transfer the thought process from that Problem to a problem like this. However when I try and follow the process I am getting stuck.
Case 1: Event $Y$ s.t. $Pr(\mathrm{only \ Yellow\ transfered})= 0$ This is because you have $\frac{^{100}C_{200}}{^{400}C_{200}} = (0/\#)=0$. It also logically makes sense, if we have to move $200$ balls from I, and there are only $100$ Yellow balls in I, then there no way for only yellow balls to be transfered.
Case 2 Event $P$ s.t. $Pr(\mathrm{different \ are\ transfered})$ This is where I am having some trouble. There can be $\{(1 \mathrm{\ yellow \ ball, 199\ blue}), (2 Y, 198 B), ..., (99Y, 101B)\}$ and I am stuck on how to figure out the number of ways that different balls can be transfered. If I know this then the $Pr(P)=\frac{\#\{\mathrm{different\ ball\ combinations}\}}{^{400}C_{200}}$.
After this i should do probability of picking yellow $={\mathrm{number\ of \ yellow}}{600}$. I should divide by $600$ because that is the total number of balls in $II$ after the transfer.
Case 3 Event $B$ s.t. $Pr(\mathrm{only blue are transferred})$ this is $\frac{^{300}C_{200}}{^{400}C_{200}} = \frac{300!/(100!200!)}{400!/(200!200!)} =$ some number, but I don't understand how to simplify this. if the numbers were larger, i think i would have an overflow problem on my calculator.
However if $Pr($only blue transfer$)=X$, then the probability of picking yellow after the transfer is $(300/600)=1/2$
Then the last step should be to add up the probabilities of each case.
You have $\frac 23$ chance to draw a ball that was originally in urn II. If you do, you have $\frac 34$ chance to get a yellow ball. You have $\frac 13$ chance to get a ball that was originally in urn I. If you do, you have $\frac 14$ chance to get a yellow ball. The overall chance of a yellow ball is then $$\frac 23 \cdot \frac 34 + \frac 13 \cdot \frac 14=\frac 7{12}$$