Question
There is a large container which will hold the following:
it is green ,green ,blue ,orang ,red ,orang ,purple ,green, purple , red
The balls are all the same size and texture. You are required to select two balls.
If both of the balls that you select are purple then you win.
How much less likely are you to win if you are required to replace the first ball? Show all of your work and explain your steps.
Answers
GPT-4o mini
To find out how much less likely you are to win when you must replace the first ball, we can calculate the probability of winning in both scenarios: with and without replacement.
Let's start by defining our situation:
**Container Composition**
- Green: 3
- Blue: 1
- Orange: 2
- Red: 2
- Purple: 2
Total number of balls = 3 (Green) + 1 (Blue) + 2 (Orange) + 2 (Red) + 2 (Purple) = 10 balls.
### Scenario 1: No replacement
1. **First Ball**: Probability of picking a purple ball = Number of purple balls / Total balls = 2/10.
2. After picking a purple ball, there is 1 purple ball left out of 9 remaining total balls.
3. **Second Ball**: Probability of picking a second purple ball = Number of purple balls left / Total balls left = 1/9.
4. **Total Probability of Winning** (both balls are purple):
\[
P(\text{Win without replacement}) = P(\text{First purple}) \times P(\text{Second purple | First purple}) = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90} = \frac{1}{45}.
\]
### Scenario 2: With replacement
1. **First Ball**: Probability of picking a purple ball = Number of purple balls / Total balls = 2/10.
2. Since we replace the first ball before drawing the second one, the total count of balls remains the same.
3. **Second Ball**: Probability of picking a purple ball again = Number of purple balls / Total balls = 2/10.
4. **Total Probability of Winning** (both balls are purple):
\[
P(\text{Win with replacement}) = P(\text{First purple}) \times P(\text{Second purple | First purple}) = \frac{2}{10} \times \frac{2}{10} = \frac{2 \times 2}{10 \times 10} = \frac{4}{100} = \frac{1}{25}.
\]
### Comparison of Probabilities
Now we can compare the two probabilities for winning:
1. Probability of winning without replacement: \( P(\text{Win without replacement}) = \frac{1}{45} \).
2. Probability of winning with replacement: \( P(\text{Win with replacement}) = \frac{1}{25} \).
Next, let's convert these fractions to a common denominator to compare them easily. The least common multiple of 45 and 25 is 225.
- Convert \( \frac{1}{45} \):
\[
\frac{1 \times 5}{45 \times 5} = \frac{5}{225}.
\]
- Convert \( \frac{1}{25} \):
\[
\frac{1 \times 9}{25 \times 9} = \frac{9}{225}.
\]
### Finding the Difference
Now we calculate how much less likely you are to win with replacement:
\[
\text{Difference} = P(\text{Win with replacement}) - P(\text{Win without replacement}) = \frac{9}{225} - \frac{5}{225} = \frac{4}{225}.
\]
### Conclusion
You are \( \frac{4}{225} \) less likely to win if you are required to replace the first ball.
Let's start by defining our situation:
**Container Composition**
- Green: 3
- Blue: 1
- Orange: 2
- Red: 2
- Purple: 2
Total number of balls = 3 (Green) + 1 (Blue) + 2 (Orange) + 2 (Red) + 2 (Purple) = 10 balls.
### Scenario 1: No replacement
1. **First Ball**: Probability of picking a purple ball = Number of purple balls / Total balls = 2/10.
2. After picking a purple ball, there is 1 purple ball left out of 9 remaining total balls.
3. **Second Ball**: Probability of picking a second purple ball = Number of purple balls left / Total balls left = 1/9.
4. **Total Probability of Winning** (both balls are purple):
\[
P(\text{Win without replacement}) = P(\text{First purple}) \times P(\text{Second purple | First purple}) = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90} = \frac{1}{45}.
\]
### Scenario 2: With replacement
1. **First Ball**: Probability of picking a purple ball = Number of purple balls / Total balls = 2/10.
2. Since we replace the first ball before drawing the second one, the total count of balls remains the same.
3. **Second Ball**: Probability of picking a purple ball again = Number of purple balls / Total balls = 2/10.
4. **Total Probability of Winning** (both balls are purple):
\[
P(\text{Win with replacement}) = P(\text{First purple}) \times P(\text{Second purple | First purple}) = \frac{2}{10} \times \frac{2}{10} = \frac{2 \times 2}{10 \times 10} = \frac{4}{100} = \frac{1}{25}.
\]
### Comparison of Probabilities
Now we can compare the two probabilities for winning:
1. Probability of winning without replacement: \( P(\text{Win without replacement}) = \frac{1}{45} \).
2. Probability of winning with replacement: \( P(\text{Win with replacement}) = \frac{1}{25} \).
Next, let's convert these fractions to a common denominator to compare them easily. The least common multiple of 45 and 25 is 225.
- Convert \( \frac{1}{45} \):
\[
\frac{1 \times 5}{45 \times 5} = \frac{5}{225}.
\]
- Convert \( \frac{1}{25} \):
\[
\frac{1 \times 9}{25 \times 9} = \frac{9}{225}.
\]
### Finding the Difference
Now we calculate how much less likely you are to win with replacement:
\[
\text{Difference} = P(\text{Win with replacement}) - P(\text{Win without replacement}) = \frac{9}{225} - \frac{5}{225} = \frac{4}{225}.
\]
### Conclusion
You are \( \frac{4}{225} \) less likely to win if you are required to replace the first ball.