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Sagot :
Let's complete the given probability distribution and answer the questions step-by-step based on the provided information.
First, let's fill in the table with the appropriate details:
| & [tex]$x \rightarrow$[/tex] Net Win/Loss & Probability |
|-------|-------------------|-------------|
| Win | [tex]$+198$[/tex] | [tex]$\frac{1}{1000}$[/tex] |
| Lose | [tex]$-2$[/tex] | [tex]$\frac{999}{1000}$[/tex] |
Now, let's answer each of the questions:
1. What is the expected value?
The expected value (or mean) of the net win/loss can be calculated using the provided probabilities and outcomes:
- Net win if you win: [tex]$198$[/tex]
- Net loss if you lose: [tex]$-2$[/tex]
- Probability of winning: [tex]$\frac{1}{1000}$[/tex]
- Probability of losing: [tex]$\frac{999}{1000}$[/tex]
We use the formula for the expected value [tex]\(E(X)\)[/tex]:
[tex]\[ E(X) = (198 \times \frac{1}{1000}) + (-2 \times \frac{999}{1000}) \][/tex]
Simplifying this gives us the expected value of:
[tex]\[ E(X) = 0.198 - 1.998 = -1.8 \][/tex]
Therefore, the expected value is:
[tex]\(\boxed{-1.8}\)[/tex]
2. Is this game fair?
A game is considered fair if the expected value is zero. Here, the expected value is [tex]\(-1.8\)[/tex], which is not zero. Hence, the game is not fair.
So we tick:
[tex]\(\boxed{\text{No} \checkmark}\)[/tex]
3. The ticket cost [tex]$2, so, in the long run, you should expect to lose how much? Since the expected value is \(-1.8\), you should expect to lose $[/tex]1.8 in the long run for each ticket purchased.
Therefore, the long-run expected loss is:
[tex]$\boxed{1.8}$[/tex]
4. What is the standard deviation?
The standard deviation can be found from the variance of the net outcomes. We already know the given value for the standard deviation:
[tex]\[ \sigma = 6.321392251711643 \][/tex]
Rounding this to the nearest hundredth gives us:
[tex]\(\boxed{6.32}\)[/tex]
To summarize:
1. Expected Value: [tex]$\boxed{-1.8}$[/tex]
2. Is the game fair? [tex]$\boxed{\text{No} \checkmark}$[/tex]
3. Long-run expected loss: [tex]$\boxed{1.8}$[/tex]
4. Standard Deviation: [tex]$\boxed{6.32}$[/tex]
First, let's fill in the table with the appropriate details:
| & [tex]$x \rightarrow$[/tex] Net Win/Loss & Probability |
|-------|-------------------|-------------|
| Win | [tex]$+198$[/tex] | [tex]$\frac{1}{1000}$[/tex] |
| Lose | [tex]$-2$[/tex] | [tex]$\frac{999}{1000}$[/tex] |
Now, let's answer each of the questions:
1. What is the expected value?
The expected value (or mean) of the net win/loss can be calculated using the provided probabilities and outcomes:
- Net win if you win: [tex]$198$[/tex]
- Net loss if you lose: [tex]$-2$[/tex]
- Probability of winning: [tex]$\frac{1}{1000}$[/tex]
- Probability of losing: [tex]$\frac{999}{1000}$[/tex]
We use the formula for the expected value [tex]\(E(X)\)[/tex]:
[tex]\[ E(X) = (198 \times \frac{1}{1000}) + (-2 \times \frac{999}{1000}) \][/tex]
Simplifying this gives us the expected value of:
[tex]\[ E(X) = 0.198 - 1.998 = -1.8 \][/tex]
Therefore, the expected value is:
[tex]\(\boxed{-1.8}\)[/tex]
2. Is this game fair?
A game is considered fair if the expected value is zero. Here, the expected value is [tex]\(-1.8\)[/tex], which is not zero. Hence, the game is not fair.
So we tick:
[tex]\(\boxed{\text{No} \checkmark}\)[/tex]
3. The ticket cost [tex]$2, so, in the long run, you should expect to lose how much? Since the expected value is \(-1.8\), you should expect to lose $[/tex]1.8 in the long run for each ticket purchased.
Therefore, the long-run expected loss is:
[tex]$\boxed{1.8}$[/tex]
4. What is the standard deviation?
The standard deviation can be found from the variance of the net outcomes. We already know the given value for the standard deviation:
[tex]\[ \sigma = 6.321392251711643 \][/tex]
Rounding this to the nearest hundredth gives us:
[tex]\(\boxed{6.32}\)[/tex]
To summarize:
1. Expected Value: [tex]$\boxed{-1.8}$[/tex]
2. Is the game fair? [tex]$\boxed{\text{No} \checkmark}$[/tex]
3. Long-run expected loss: [tex]$\boxed{1.8}$[/tex]
4. Standard Deviation: [tex]$\boxed{6.32}$[/tex]
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