Question

A bag contains 3 red, 2 yellow, and 3 green marbles. You reach in and draw out three marbles without replacement.

What is the probability that the first two marbles will be green and the third will be yellow?

Answer 1

To solve this problem, we need to calculate the probability of drawing two green marbles followed by a yellow marble from the bag without replacement.

Let's analyze the marble distribution in the bag:
- Red marbles: 3
- Yellow marbles: 2
- Green marbles: 3

This gives us a total of \(3 + 2 + 3 = 8\) marbles in the bag.

We want to find the probability of drawing the first two marbles as green, followed by one yellow.

1. **Probability of drawing the first green marble:**
- There are 3 green marbles out of 8 total marbles. Therefore, the probability of drawing a green marble first is:
\[
P(\text{first green}) = \frac{3}{8}
\]

2. **Probability of drawing the second green marble:**
- After drawing the first green marble, there are now 2 green marbles left out of 7 total marbles remaining. The probability of drawing a second green marble is:
\[
P(\text{second green} \mid \text{first green}) = \frac{2}{7}
\]

3. **Probability of drawing a yellow marble third:**
- After drawing two green marbles, there remain 6 marbles in total: 3 red, 2 yellow, and 1 green. The probability of drawing a yellow marble next is:
\[
P(\text{yellow} \mid \text{first green, second green}) = \frac{2}{6} = \frac{1}{3}
\]

Now, we multiply these probabilities together to get the total probability of drawing this specific sequence of marbles:

\[
P(\text{first two green, third yellow}) = P(\text{first green}) \times P(\text{second green} \mid \text{first green}) \times P(\text{yellow} \mid \text{first green, second green})
\]
\[
= \frac{3}{8} \times \frac{2}{7} \times \frac{1}{3}
\]

Calculating this step-by-step:

1. Multiply the first two fractions:
\[
\frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}
\]

2. Now multiply this result by the yellow probability:
\[
\frac{3}{28} \times \frac{1}{3} = \frac{3}{84} = \frac{1}{28}
\]

Thus, the probability that the first two marbles drawn are green and the third marble is yellow is:

\[
\boxed{\frac{1}{28}}
\]