The image shows a probability question asking for $P(B \text{ and } B)$, which means finding the probability of drawing a 'B' tile in two successive draws from a bag. The tiles in the bag are as follows:

• 4 A's
• 2 B's
• 2 C's

The tiles are replaced after the first draw, so the draws are independent.

To solve this, we use the following steps:

1. Probability of drawing a B on the first draw (P(B)):
   Since there are 2 B's out of 8 tiles (4 A's + 2 B's + 2 C's),
   $P(B) = \frac{2}{8} = \frac{1}{4}$

2. Probability of drawing a B on the second draw (P(B)):
   Since the tiles are replaced, the total number of tiles is still 8, and there are still 2 B's. So,
   $P(B) = \frac{2}{8} = \frac{1}{4}$

3. Finding the probability of drawing B on both draws (P(B and B)):
   Since the events are independent (due to replacement), we multiply the probabilities of each draw: $P(B \text{ and } B) = P(B) \times P(B) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

Thus, the answer is $\frac{1}{16}$.

Let me know if you need further explanations or details!

Here are some related questions for further understanding:

1. How would the probability change if the tiles were not replaced after each draw?
2. What is the probability of drawing at least one 'B' in two draws?
3. How would the answer change if there were 3 B's and 3 C's instead of 2 B's and 2 C's?
4. Can you calculate the probability of drawing two 'A's in the same scenario?
5. How does the probability change if the number of tiles increases but the ratio of B's remains the same?

Tip: In probability, when events are independent (like drawing with replacement), the probability of both events happening is found by multiplying the probabilities of each event separately.