An urn contains three red marbles and four blue marbles. One marble is selected at random, its color is noted, and it is returned to the urn. Another marble is then selected. What is the probability the second marble drawn is red? 117 N/7 337 36
An urn contains three red marbles and four blue marbles. One marble is selected at random, its color is noted, and it is returned to the urn. Another marble is then selected. What is the probability the second marble drawn is red? 117 N/7 337 36
Chapter1: Financial Statements And Business Decisions
Section: Chapter Questions
Problem 1Q
Related questions
Question
![**Question:**
An urn contains three red marbles and four blue marbles. One marble is selected at random, its color is noted, and it is returned to the urn. Another marble is then selected. What is the probability the second marble drawn is red?
**Options:**
- \( \frac{1}{7} \)
- \( \frac{2}{7} \)
- \( \frac{3}{7} \)
- \( \frac{3}{6} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F064aec2b-96bf-4be0-a1f2-689bafa9211a%2F163353b2-39a9-4218-91b8-f5a170069350%2Ftv4ypgj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
An urn contains three red marbles and four blue marbles. One marble is selected at random, its color is noted, and it is returned to the urn. Another marble is then selected. What is the probability the second marble drawn is red?
**Options:**
- \( \frac{1}{7} \)
- \( \frac{2}{7} \)
- \( \frac{3}{7} \)
- \( \frac{3}{6} \)
![**Question**
The probability of drawing a red marble from a sack of marbles is \( \frac{2}{5} \). Which set of marbles could the sack contain?
- ○ 2 red marbles and 5 green marbles
- ○ 4 red marbles and 6 green marbles
- ○ 6 red marbles and 15 green marbles
- ○ 2 red marbles, 1 blue marble, and 4 white marbles
**Explanation**
To solve this problem, we need to find a set where the ratio of red marbles to the total number of marbles equals \( \frac{2}{5} \).
1. **Option 1**: 2 red marbles and 5 green marbles
- Total marbles = 2 + 5 = 7
- Probability = \( \frac{2}{7} \)
2. **Option 2**: 4 red marbles and 6 green marbles
- Total marbles = 4 + 6 = 10
- Probability = \( \frac{4}{10} = \frac{2}{5} \) (Correct Choice)
3. **Option 3**: 6 red marbles and 15 green marbles
- Total marbles = 6 + 15 = 21
- Probability = \( \frac{6}{21} = \frac{2}{7} \)
4. **Option 4**: 2 red marbles, 1 blue marble, and 4 white marbles
- Total marbles = 2 + 1 + 4 = 7
- Probability = \( \frac{2}{7} \)
The correct answer is the second option: 4 red marbles and 6 green marbles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F064aec2b-96bf-4be0-a1f2-689bafa9211a%2F163353b2-39a9-4218-91b8-f5a170069350%2Fu9t1ccx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question**
The probability of drawing a red marble from a sack of marbles is \( \frac{2}{5} \). Which set of marbles could the sack contain?
- ○ 2 red marbles and 5 green marbles
- ○ 4 red marbles and 6 green marbles
- ○ 6 red marbles and 15 green marbles
- ○ 2 red marbles, 1 blue marble, and 4 white marbles
**Explanation**
To solve this problem, we need to find a set where the ratio of red marbles to the total number of marbles equals \( \frac{2}{5} \).
1. **Option 1**: 2 red marbles and 5 green marbles
- Total marbles = 2 + 5 = 7
- Probability = \( \frac{2}{7} \)
2. **Option 2**: 4 red marbles and 6 green marbles
- Total marbles = 4 + 6 = 10
- Probability = \( \frac{4}{10} = \frac{2}{5} \) (Correct Choice)
3. **Option 3**: 6 red marbles and 15 green marbles
- Total marbles = 6 + 15 = 21
- Probability = \( \frac{6}{21} = \frac{2}{7} \)
4. **Option 4**: 2 red marbles, 1 blue marble, and 4 white marbles
- Total marbles = 2 + 1 + 4 = 7
- Probability = \( \frac{2}{7} \)
The correct answer is the second option: 4 red marbles and 6 green marbles.
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