Question
There is a large container which will hold the following:
pasted-image-1708016034925_cdee86b2-b48d-4798-a1bd-12d3e0249793
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 determine how much less likely you are to win by replacing the first ball you select, we need to analyze two scenarios: selecting two balls without replacement and selecting two balls with replacement.
**1. Selecting Two Balls Without Replacement:**
Let's assume that the container holds a total of `N` balls, of which `M` are purple and `N - M` are of other colors.
- **Probability of Drawing Two Purple Balls Without Replacement:**
- For the first ball, the probability of drawing a purple ball is \( \frac{M}{N} \).
- After drawing one purple ball, there are now \( M - 1 \) purple balls left and only \( N - 1 \) balls total. So, the probability of drawing a second purple ball is \( \frac{M - 1}{N - 1} \).
- Therefore, the probability of drawing two purple balls without replacement is:
\[
P(\text{win without replacement}) = \frac{M}{N} \times \frac{M - 1}{N - 1}
\]
**2. Selecting Two Balls With Replacement:**
Now, if we replace the first ball after drawing it, both draws are independent.
- **Probability of Drawing Two Purple Balls With Replacement:**
- The probability of drawing a purple ball remains \( \frac{M}{N} \) for both the first and second draws.
- Thus, the probability of drawing two purple balls with replacement is:
\[
P(\text{win with replacement}) = \frac{M}{N} \times \frac{M}{N} = \left( \frac{M}{N} \right)^2
\]
**3. Comparing the Two Probabilities:**
Now, we can compare the two probabilities to see how much less likely you are to win when replacing the first ball.
- **Without Replacement:**
\[
P(\text{win without replacement}) = \frac{M}{N} \times \frac{M - 1}{N - 1}
\]
- **With Replacement:**
\[
P(\text{win with replacement}) = \left( \frac{M}{N} \right)^2
\]
**4. How Much Less Likely:**
To find how much less likely you are to win if you have to replace the first ball, we can compare the two probabilities. We want to calculate:
\[
\Delta P = P(\text{win without replacement}) - P(\text{win with replacement})
\]
**5. Calculation Example:**
If, for example, \( N = 10 \) and \( M = 5 \):
- Calculation for without replacement:
\[
P(\text{win without replacement}) = \frac{5}{10} \times \frac{4}{9} = \frac{5 \times 4}{10 \times 9} = \frac{20}{90} = \frac{2}{9}
\]
- Calculation for with replacement:
\[
P(\text{win with replacement}) = \left( \frac{5}{10} \right)^2 = \frac{25}{100} = \frac{1}{4}
\]
Now we can compare these values to find out how much less likely you are to win:
**6. Final Calculation:**
Finding a common denominator to compare \( \frac{2}{9} \) and \( \frac{1}{4} \):
\[
\frac{2}{9} \approx 0.2222
\]
\[
\frac{1}{4} = 0.25
\]
To determine difference:
\[
\Delta P = \frac{2}{9} - \frac{1}{4} = \frac{8}{36} - \frac{9}{36} = -\frac{1}{36}
\]
So, you are less likely to win when replacing the ball than if you did not replace it. The magnitude of this difference, in the context of probabilities, is what we derive from the specific values of \( M \) and \( N \).
Thus, this shows how to assess the likelihood of winning based on whether or not the selection of the balls is with or without replacement.
**1. Selecting Two Balls Without Replacement:**
Let's assume that the container holds a total of `N` balls, of which `M` are purple and `N - M` are of other colors.
- **Probability of Drawing Two Purple Balls Without Replacement:**
- For the first ball, the probability of drawing a purple ball is \( \frac{M}{N} \).
- After drawing one purple ball, there are now \( M - 1 \) purple balls left and only \( N - 1 \) balls total. So, the probability of drawing a second purple ball is \( \frac{M - 1}{N - 1} \).
- Therefore, the probability of drawing two purple balls without replacement is:
\[
P(\text{win without replacement}) = \frac{M}{N} \times \frac{M - 1}{N - 1}
\]
**2. Selecting Two Balls With Replacement:**
Now, if we replace the first ball after drawing it, both draws are independent.
- **Probability of Drawing Two Purple Balls With Replacement:**
- The probability of drawing a purple ball remains \( \frac{M}{N} \) for both the first and second draws.
- Thus, the probability of drawing two purple balls with replacement is:
\[
P(\text{win with replacement}) = \frac{M}{N} \times \frac{M}{N} = \left( \frac{M}{N} \right)^2
\]
**3. Comparing the Two Probabilities:**
Now, we can compare the two probabilities to see how much less likely you are to win when replacing the first ball.
- **Without Replacement:**
\[
P(\text{win without replacement}) = \frac{M}{N} \times \frac{M - 1}{N - 1}
\]
- **With Replacement:**
\[
P(\text{win with replacement}) = \left( \frac{M}{N} \right)^2
\]
**4. How Much Less Likely:**
To find how much less likely you are to win if you have to replace the first ball, we can compare the two probabilities. We want to calculate:
\[
\Delta P = P(\text{win without replacement}) - P(\text{win with replacement})
\]
**5. Calculation Example:**
If, for example, \( N = 10 \) and \( M = 5 \):
- Calculation for without replacement:
\[
P(\text{win without replacement}) = \frac{5}{10} \times \frac{4}{9} = \frac{5 \times 4}{10 \times 9} = \frac{20}{90} = \frac{2}{9}
\]
- Calculation for with replacement:
\[
P(\text{win with replacement}) = \left( \frac{5}{10} \right)^2 = \frac{25}{100} = \frac{1}{4}
\]
Now we can compare these values to find out how much less likely you are to win:
**6. Final Calculation:**
Finding a common denominator to compare \( \frac{2}{9} \) and \( \frac{1}{4} \):
\[
\frac{2}{9} \approx 0.2222
\]
\[
\frac{1}{4} = 0.25
\]
To determine difference:
\[
\Delta P = \frac{2}{9} - \frac{1}{4} = \frac{8}{36} - \frac{9}{36} = -\frac{1}{36}
\]
So, you are less likely to win when replacing the ball than if you did not replace it. The magnitude of this difference, in the context of probabilities, is what we derive from the specific values of \( M \) and \( N \).
Thus, this shows how to assess the likelihood of winning based on whether or not the selection of the balls is with or without replacement.