To solve the problem involving the geometric distribution for the daily number lottery, we can apply the formulas for the mean and variance.

Given:

• Probability of winning (p) = $\frac{1}{1000}$
• Probability of losing (q) = $1 - p = \frac{999}{1000}$

(a) Mean, Variance, and Standard Deviation

1. Mean (μ): $\mu = \frac{1}{p} = \frac{1}{\frac{1}{1000}} = 1000$

2. Variance (σ²): $\sigma^2 = \frac{q}{p^2} = \frac{\frac{999}{1000}}{\left(\frac{1}{1000}\right)^2} = \frac{999 \cdot 1000^2}{1000} = 999000$

3. Standard Deviation (σ): $\sigma = \sqrt{\sigma^2} = \sqrt{999000} \approx 999.5$

Summary of Results:

• Mean (μ): 1000
• Variance (σ²): 999000
• Standard Deviation (σ): Approximately 999.5

(b) Expected Plays and Profit/Loss

• Expected number of plays before winning: 1000

To evaluate whether you would expect to make or lose money:

1. Cost per play: $1
2. Expected cost before winning: $1000 \times 1 = 1000$ dollars.
3. Winning payout: $800

Net Expected Value: $\text{Net Expected Value} = \text{Winning payout} - \text{Expected cost} = 800 - 1000 = -200$

Conclusion:

You would expect to lose $200 on average for every 1000 plays.

Questions for Further Exploration:

1. How would the expected number of plays change if the winning probability increased?
2. What would be the variance if the lottery had a different payout structure?
3. How does the expected value change with multiple tickets bought per draw?
4. What are the implications of playing this lottery over the long term?
5. How could one optimize their strategy in lotteries with different rules?

Tip: Always consider the expected value and variance when evaluating games of chance to make informed decisions. Would you like more details on any part of this?