Question

What is the probability of drawing a tile with X and then a tile with Z without replacement from a bag containing 6 X tiles, 8 Y tiles and 3 Z tiles?

Answer 1

To find the probability of drawing a tile with X and then a tile with Z without replacement from the given bag, we start with the total number of tiles in the bag.

1. **Calculate total tiles:**
- X tiles = 6
- Y tiles = 8
- Z tiles = 3
- Total tiles = 6 + 8 + 3 = 17

2. **Finding the probability of first drawing an X tile:**
The probability \( P(X) \) of drawing an X tile first is:
\[
P(X) = \frac{\text{Number of X tiles}}{\text{Total number of tiles}} = \frac{6}{17}
\]

3. **Change the total after drawing an X tile:**
After drawing an X tile, there are now:
- Total tiles left = 17 - 1 = 16
- Z tiles remain = 3 (unchanged)

4. **Finding the probability of drawing a Z tile after drawing an X tile:**
The probability \( P(Z|X) \) of then drawing a Z tile is:
\[
P(Z|X) = \frac{\text{Number of Z tiles}}{\text{Total number of tiles left}} = \frac{3}{16}
\]

5. **Finding the combined probability:**
To find the combined probability of both events (drawing an X tile first and then a Z tile), we multiply the probabilities:
\[
P(X \text{ and then } Z) = P(X) \times P(Z|X) = \frac{6}{17} \times \frac{3}{16} = \frac{18}{272}
\]

6. **Simplifying the result:**
Dividing the numerator and the denominator by 2:
\[
\frac{18}{272} = \frac{9}{136}
\]

Thus, the probability of drawing a tile with X and then a tile with Z without replacement is \( \frac{9}{136} \).