Bag 1: 6b,5w
Bag 2: 5b,6w
3 balls are transferred from bag 1 to bag 2 at random.
2 balls are drawn from bag 2. What's the probability to take 2 balls of different colours ?
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you need to sum up the probabilities of all the cases of the 3 ball transfer
$P[\text{2 balls from bag 2 were different color}] = \sum \limits_{\vec{t}}P[\text{2 balls from bag 2 were different color} | \vec{t}]P[\vec{t}]\\
\text{where }\vec{t} \text{ is the combination of balls that were transferred from bag 1}$
for example for $\vec{t}=(3,0) =\text{ (3 white, 0 black)}$
$P[\text{2 different colors from bag 2}|(3,0)]P[(3,0)] = \dfrac{\dbinom{5}{1}\dbinom{9}{1}}{\dbinom{14}{2}} \cdot \dfrac{\dbinom{6}{0}\dbinom{5}{3}}{\dbinom{11}{3}} $
repeat this for all the combinations of the 3 black and white balls and sum them all up.
Below is the summation on the conditional probabilities for the number of black balls moved from I to II being $k=0,\cdots, 3$
Example: Say there two Black Balls & one White moved. Lets call that event $\mathcal{A}$ and the event that the second draw of two balls have the same colour is $\mathcal{B}$.
The probabilities are: $\mathscr{P}(\mathcal{A})=\dfrac{\dbinom{6}{2} \dbinom{5}{1}}{\dbinom{11}{3}}$ and for next draw, $\mathscr{P}(\mathcal{B})=\dfrac{\dbinom{7}{2}+ \dbinom{7}{2}}{\dbinom{14}{2}}$
$\displaystyle{\large\sum\limits_{k = 0}^3} \left\{\dfrac{\dbinom{5+k}{2}+\dbinom{9-k}{2}}{\dbinom{14}{2}}\right\}\left[\dfrac{\dbinom{6}{k}\dbinom{5}{3-k}}{\dbinom{11}{3}}\right] $