Probability of getting white balls from urns QUESTION: An urn $A$ contains $3$ white balls and $4$ black balls. $2$ balls are drawn from urn $A$ and without seeing the colour and we introduce them in urn $B$ (which was empty). Later, a ball is drawn randomly from each urn.
a) Find out the probability that the ball drawn from urn $A$ is white.
b) Find out the probability that the two balls drawn are white.
BELOW IS MY SOLUTION. Is it correct?
a) Let $w$ the event of drawing a white ball and $b$ the event of drawing a black ball. The probabilities for the events of two drawings from urn $A$ are:
$\begin{array}{lll}
P(\{ ww \}) &=& \displaystyle \frac{3}{7} \cdot \frac{2}{6} = \frac{1}{7}, \quad P(\{ wb \}) = \displaystyle \frac{3}{7} \cdot \frac{4}{6} = \frac{2}{7}, \\
P(\{ bw \}) &=& \displaystyle \frac{4}{7} \cdot \frac{3}{6} = \frac{2}{7}, \quad P(\{ bb \}) = \displaystyle \frac{4}{7} \cdot \frac{3}{6} = \frac{2}{7} \\
\end{array}$
Let $W_A$ the event of drawing a white ball from urn $A$ after $2$ drawings. By total probability theorem:
$\begin{array}{lll}
P\left( W_A \right) & = & \displaystyle P\left(  \{ ww \} \right) \cdot P\left(W_A | \{ ww \} \right) + P\left(  \{ wb \} \right) \cdot P\left(W_A | \{ bn \} \right) + \\
&& P\left(  \{ bw \} \right) \cdot P\left(W_A | \{ bw \} \right) + P\left(  \{ bb \} \right) \cdot P\left(W_A | \{ bb \} \right)  \\
& = & \displaystyle \frac{1}{7} \cdot \frac{1}{5} + \frac{2}{7} \cdot \frac{2}{5} + \frac{2}{7} \cdot \frac{2}{5} + \frac{2}{7} \cdot \frac{3}{5} = \frac{1+4+4+6}{35}=\frac{15}{35} = \frac{3}{7} \\
\end{array}$
Consequently, the probability that the ball drawn from urn $A$ is white is $\displaystyle \frac{3}{7}$.
b) Let be the following events:
$\begin{array}{lll}
W_B & = & \{ \text{Get white ball from urn $B$ after $2$ drawings} \} \\
W_A \cap W_B & = & \{ \text{Get $2$ white balls after $2$ drawings} \} \\
\end{array}$
By total probability theorem:
$\begin{array}{lll}
P\left( W_A \cap W_B \right) & = & \displaystyle P\left( \{ bb \} \right) \cdot P\left(W_A \cap W_B | \{ bb \} \right) + P\left(  \{ bn \} \right) \cdot P\left(W_A \cap W_B | \{ bn \} \right) + \\
&& P\left(  \{ nb \} \right) \cdot P\left(W_A \cap W_B | \{ nb \} \right) + P\left(  \{ nn \} \right) \cdot P\left(W_A \cap W_B | \{ nn \} \right)  \\
& = & \displaystyle \frac{1}{7} \cdot \frac{1}{5} \cdot \frac{2}{2} + \frac{2}{7} \cdot \frac{2}{5} \cdot \frac{1}{2} + \frac{2}{7} \cdot \frac{2}{5} \cdot \frac{1}{2} + \frac{2}{7} \cdot \frac{3}{5} \cdot \frac{0}{2} = \frac{1+2+2}{35}=\frac{5}{35} = \frac{1}{7} \\
\end{array}$
The probability that the two balls drawn are white is $\displaystyle \frac{1}{7}$.

Given the solutions, we are hinted that the procedure of choosing $1$ or $2$ white balls does not affect the answer.
$$ \frac {3 \choose 1} {7 \choose 1} = {3 \over 7} $$
$$ \frac {3 \choose 2} {7 \choose 2} = {1 \over 7} $$
This makes sense.
Let's think to a deck with red an black cards. It does not matter how much we shuffle, cut or re-shuffle before we randomly turn two cards down side up and reveal the color.
