Question
Consider two bags containing 20 marbles. In each bag, there are 5 red, 3 green, 2 yellow with stripes, 7 blue, and 3 green with stripes. What is the probability of selecting a green marble from the first bag and a striped marble from the second bag? Write your answer as a reduced fraction.(1 point)
Answers
GPT-4o mini
To find the probability of selecting a green marble from the first bag and a striped marble from the second bag, we will calculate each probability separately and then multiply them together.
1. **Probability of selecting a green marble from the first bag:**
- In the first bag, there are a total of 20 marbles, consisting of:
- 5 red
- 3 green
- 2 yellow with stripes
- 7 blue
- 3 green with stripes
- The total number of green marbles (both plain green and green with stripes) is \(3 + 3 = 6\).
The probability of selecting a green marble from the first bag is:
\[
P(\text{green from first bag}) = \frac{\text{number of green marbles}}{\text{total number of marbles}} = \frac{6}{20} = \frac{3}{10}
\]
2. **Probability of selecting a striped marble from the second bag:**
- Similar to the first bag, the second bag also contains a total of 20 marbles consisting of the same distribution. The striped marbles consist of:
- 2 yellow with stripes
- 3 green with stripes
- The total number of striped marbles is \(2 + 3 = 5\).
The probability of selecting a striped marble from the second bag is:
\[
P(\text{striped from second bag}) = \frac{\text{number of striped marbles}}{\text{total number of marbles}} = \frac{5}{20} = \frac{1}{4}
\]
3. **Combined probability:**
To find the probability of both events happening (selecting a green marble from the first bag and a striped marble from the second bag), we multiply the two probabilities together:
\[
P(\text{green from first bag and striped from second bag}) = P(\text{green from first bag}) \times P(\text{striped from second bag}) = \frac{3}{10} \times \frac{1}{4} = \frac{3 \times 1}{10 \times 4} = \frac{3}{40}
\]
Thus, the probability of selecting a green marble from the first bag and a striped marble from the second bag is \(\frac{3}{40}\).
The final answer is:
\[
\boxed{\frac{3}{40}}
\]
1. **Probability of selecting a green marble from the first bag:**
- In the first bag, there are a total of 20 marbles, consisting of:
- 5 red
- 3 green
- 2 yellow with stripes
- 7 blue
- 3 green with stripes
- The total number of green marbles (both plain green and green with stripes) is \(3 + 3 = 6\).
The probability of selecting a green marble from the first bag is:
\[
P(\text{green from first bag}) = \frac{\text{number of green marbles}}{\text{total number of marbles}} = \frac{6}{20} = \frac{3}{10}
\]
2. **Probability of selecting a striped marble from the second bag:**
- Similar to the first bag, the second bag also contains a total of 20 marbles consisting of the same distribution. The striped marbles consist of:
- 2 yellow with stripes
- 3 green with stripes
- The total number of striped marbles is \(2 + 3 = 5\).
The probability of selecting a striped marble from the second bag is:
\[
P(\text{striped from second bag}) = \frac{\text{number of striped marbles}}{\text{total number of marbles}} = \frac{5}{20} = \frac{1}{4}
\]
3. **Combined probability:**
To find the probability of both events happening (selecting a green marble from the first bag and a striped marble from the second bag), we multiply the two probabilities together:
\[
P(\text{green from first bag and striped from second bag}) = P(\text{green from first bag}) \times P(\text{striped from second bag}) = \frac{3}{10} \times \frac{1}{4} = \frac{3 \times 1}{10 \times 4} = \frac{3}{40}
\]
Thus, the probability of selecting a green marble from the first bag and a striped marble from the second bag is \(\frac{3}{40}\).
The final answer is:
\[
\boxed{\frac{3}{40}}
\]