Urn vs Dice table| No. on die's face | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|
| Urn Selected | B | A | B | A | B | A |
|---|
Additionally, Urn A has labels $\left \{\underbrace{10}_\text{3 balls}, \underbrace{0}_\text{1 ball} \right \}$ and Urn B has labels $\left \{\underbrace{5}_\text{2 balls}, \underbrace{1}_\text{6 balls} \right \}$ and $P(\text{odd face})=P(\text{even face})=\frac{1}{2}$
Let $B10, B0, \cdots$ denote the event of the listed labelled ball being drawn.
Probability of drawing balls = $\dfrac{\text{required balls}}{\text{total no. of balls in that urn}}$
So,
$$P(B10)= P(\text{even face}) \times \frac{3}{4} = \frac{1}{2}\times \frac{3}{4}=\frac{3}{8}$$
$$P(B0)= P(\text{even face}) \times \frac{1}{4} = \frac{1}{8}$$
$$P(B5)= P(\text{odd face}) \times \frac{2}{8} = \frac{1}{8}$$
$$P(B1)= P(\text{odd face}) \times \frac{6}{8} = \frac{3}{8}$$
Hence the expectation would be:
$$\boxed{\mathbb{E}(\text{die face})=\dfrac{1}{8} \left [ (10 \times 3) + 0 + 5 + 3 \right ] = \frac{38}{8} = 4.75}$$
