Question
There is a large container which hold the following: 2 green balls, 1 yellow, 2 red, 2 orange, 1 blue, 2 purple.
The balls are the same size and texture. You are required to select two balls. If both of the balls you select are purple then you win.
How much less likely are you to win if you are required to replace the first ball?
The balls are the same size and texture. You are required to select two balls. If both of the balls you select are purple then you win.
How much less likely are you to win if you are required to replace the first ball?
Answers
GPT-4o mini
To solve the problem, we first need to understand the probabilities involved in selecting the balls.
### Without Replacement
1. **Total Number of Balls**:
- Green: 2
- Yellow: 1
- Red: 2
- Orange: 2
- Blue: 1
- Purple: 2
- **Total**: \(2 + 1 + 2 + 2 + 1 + 2 = 10\) balls.
2. **Winning Condition**: Select 2 purple balls.
3. **Calculating Probability (Without Replacement)**:
- The probability of selecting a purple ball first: \(\frac{2}{10}\).
- Once one purple ball is selected, only 1 purple ball remains and 9 balls total left.
- The probability of selecting the second purple ball: \(\frac{1}{9}\).
Therefore, the probability of selecting 2 purple balls without replacement is:
\[
P(\text{win without replacement}) = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90} = \frac{1}{45}.
\]
### With Replacement
1. **Calculating Probability (With Replacement)**:
- After selecting the first ball, since we replace it, the total remains 10 balls with 2 purple.
- Probability of selecting a purple ball: \(\frac{2}{10} = \frac{1}{5}\) for both the first and second selections.
Therefore, the probability of selecting 2 purple balls with replacement is:
\[
P(\text{win with replacement}) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}.
\]
### Comparing the Probabilities
Now we need to find out how much less likely you are to win when you are required to replace the first ball:
- Probability without replacement: \(\frac{1}{45}\)
- Probability with replacement: \(\frac{1}{25}\)
To compare these probabilities, we can convert them to a common denominator. The least common multiple of 45 and 25 is 225.
1. Convert \(\frac{1}{45}\) to the common denominator:
\[
\frac{1}{45} = \frac{5}{225}.
\]
2. Convert \(\frac{1}{25}\) to the common denominator:
\[
\frac{1}{25} = \frac{9}{225}.
\]
### Calculate the Difference
Now we calculate how much less likely the probability of winning is with replacement compared to without replacement:
\[
\text{Difference} = P(\text{win with replacement}) - P(\text{win without replacement}) = \frac{9}{225} - \frac{5}{225} = \frac{4}{225}.
\]
### Conclusion
Thus, you are less likely to win by \(\frac{4}{225}\) when you are required to replace the first ball as compared to when you do not replace it.
### Without Replacement
1. **Total Number of Balls**:
- Green: 2
- Yellow: 1
- Red: 2
- Orange: 2
- Blue: 1
- Purple: 2
- **Total**: \(2 + 1 + 2 + 2 + 1 + 2 = 10\) balls.
2. **Winning Condition**: Select 2 purple balls.
3. **Calculating Probability (Without Replacement)**:
- The probability of selecting a purple ball first: \(\frac{2}{10}\).
- Once one purple ball is selected, only 1 purple ball remains and 9 balls total left.
- The probability of selecting the second purple ball: \(\frac{1}{9}\).
Therefore, the probability of selecting 2 purple balls without replacement is:
\[
P(\text{win without replacement}) = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90} = \frac{1}{45}.
\]
### With Replacement
1. **Calculating Probability (With Replacement)**:
- After selecting the first ball, since we replace it, the total remains 10 balls with 2 purple.
- Probability of selecting a purple ball: \(\frac{2}{10} = \frac{1}{5}\) for both the first and second selections.
Therefore, the probability of selecting 2 purple balls with replacement is:
\[
P(\text{win with replacement}) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}.
\]
### Comparing the Probabilities
Now we need to find out how much less likely you are to win when you are required to replace the first ball:
- Probability without replacement: \(\frac{1}{45}\)
- Probability with replacement: \(\frac{1}{25}\)
To compare these probabilities, we can convert them to a common denominator. The least common multiple of 45 and 25 is 225.
1. Convert \(\frac{1}{45}\) to the common denominator:
\[
\frac{1}{45} = \frac{5}{225}.
\]
2. Convert \(\frac{1}{25}\) to the common denominator:
\[
\frac{1}{25} = \frac{9}{225}.
\]
### Calculate the Difference
Now we calculate how much less likely the probability of winning is with replacement compared to without replacement:
\[
\text{Difference} = P(\text{win with replacement}) - P(\text{win without replacement}) = \frac{9}{225} - \frac{5}{225} = \frac{4}{225}.
\]
### Conclusion
Thus, you are less likely to win by \(\frac{4}{225}\) when you are required to replace the first ball as compared to when you do not replace it.