Question


There is a large container which will hold the following:


it is green,green,blue,orang,red,orang,blue,green,blue,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 determine how much less likely you are to win if you are required to replace the first ball when selecting two balls from the container, we first need to examine the composition of the balls and the probabilities involved.

### Step 1: Identify the Ball Colors and Their Counts
The colors of the balls in the container are:
- Green: 3
- Blue: 3
- Orange: 2
- Red: 2

The total number of balls is:
\[
3 \text{ (green)} + 3 \text{ (blue)} + 2 \text{ (orange)} + 2 \text{ (red)} = 10 \text{ balls}
\]

### Step 2: Calculate the Probability of Winning without Replacement
Winning requires drawing two purple balls. However, there are no purple balls in the container (as identified from the colors present). Therefore, the probability of winning is straightforward:

- The number of purple balls = 0

When selecting two balls without replacement, the probability of selecting two purple balls is:
\[
P(\text{two purple without replacement}) = \frac{0}{10} \times \frac{0}{9} = 0
\]

### Step 3: Calculate the Probability of Winning with Replacement
Now, let's consider the case where we replace the first ball drawn. The probability of winning with replacement is also calculated as follows:

When the first ball is drawn and replaced, the total number of balls remains constant at 10, and the same initial chance applies for the second draw.

- The probability of drawing the first purple ball is still 0.
- The probability of drawing the second purple ball (after replacing the first) is also 0.

Thus, the probability of drawing two purple balls with replacement is:
\[
P(\text{two purple with replacement}) = \frac{0}{10} \times \frac{0}{10} = 0
\]

### Step 4: Compare the Probabilities
In both scenarios—drawing without replacement and with replacement—the probability of winning remains 0 since there are no purple balls.

### Conclusion
Since the probability of winning in both cases is the same (0), you are not less likely to win with replacement:

\[
P(\text{without replacement}) = 0, \quad P(\text{with replacement}) = 0
\]

Thus, the answer to “how much less likely are you to win if you are required to replace the first ball?” is:
\[
\text{You are not less likely to win. Probability change} = 0.
\]

Therefore, because there are no purple balls, replacement doesn't affect the probability of winning, and both situations yield a win probability of zero.