Question

Suppose that two M&M's are drawn from a bag of M&M's that contains 3 green M&M's, 8 brown M&M's, 8 red M&M's and 9 blue M&M's. The first M&M is not replaced before the second M&M is drawn. What is the probability of picking a red M&M first and then a blue M&M?

Answer 1

To find the probability of drawing a red M&M first and then a blue M&M without replacement, we can follow these steps:

1. **Determine the total number of M&M's** in the bag:
- Green M&M's: 3
- Brown M&M's: 8
- Red M&M's: 8
- Blue M&M's: 9

Total M&M's = 3 + 8 + 8 + 9 = 28

2. **Calculate the probability of drawing a red M&M first**:
- The number of red M&M's = 8
- Total M&M's = 28

The probability of drawing a red M&M first:
\[
P(\text{Red first}) = \frac{8}{28} = \frac{2}{7}
\]

3. **Calculate the probability of drawing a blue M&M second** after drawing a red M&M:
- After drawing a red M&M, the total number of M&M's left is 27 (28 - 1).
- The number of blue M&M's remains the same at 9.

The probability of drawing a blue M&M second:
\[
P(\text{Blue second | Red first}) = \frac{9}{27} = \frac{1}{3}
\]

4. **Calculate the combined probability of both events**:
\[
P(\text{Red first and Blue second}) = P(\text{Red first}) \times P(\text{Blue second | Red first}) = \frac{2}{7} \times \frac{1}{3} = \frac{2}{21}
\]

Thus, the probability of picking a red M&M first and then a blue M&M is \( \frac{2}{21} \).