Question

Mr. Gray's class is conducting an experiment to find the probability of pulling certain colors from a bag of 25 marbles. If 6 are purple, 7 are yellow, 2 are green, and the rest are black, what is the probability of drawing a green and black if the marbles are not replaced after they are picked?
A.

B.

C.

D.

Answer 1

To find the probability of drawing a green marble and then a black marble from the bag without replacement, we first need to determine the total number of marbles and the number of each color:

- Total marbles = 25
- Purple marbles = 6
- Yellow marbles = 7
- Green marbles = 2
- Black marbles = 25 - (6 + 7 + 2) = 25 - 15 = 10

Now we can calculate the probability:

1. **Probability of drawing a green marble first:**
\[
P(\text{Green}) = \frac{\text{Number of green marbles}}{\text{Total number of marbles}} = \frac{2}{25}
\]

2. **After drawing one green marble, the total number of marbles decreases to 24, and the number of black marbles remains 10.**

**Probability of drawing a black marble second:**
\[
P(\text{Black | Green drawn}) = \frac{\text{Number of black marbles}}{\text{Total number of marbles remaining}} = \frac{10}{24} = \frac{5}{12}
\]

3. **Now, we can find the combined probability of both events happening:**
\[
P(\text{Green then Black}) = P(\text{Green}) \times P(\text{Black | Green drawn}) = \frac{2}{25} \times \frac{5}{12}
\]

\[
P(\text{Green then Black}) = \frac{2 \times 5}{25 \times 12} = \frac{10}{300} = \frac{1}{30}
\]

Thus, the probability of drawing a green marble first and a black marble second without replacement is \(\frac{1}{30}\).

If you're provided with options A, B, C, and D, you can choose the one that corresponds to \(\frac{1}{30}\).