To solve this, we can treat this situation as a hypergeometric distribution, since the tickets are not replaced once selected.

Parameters:

• Total number of tickets (population size): 500
• Number of winning tickets (successes): 50
• Number of tickets bought by one person (sample size): 20
• X: The number of winning tickets bought by the person.

Hypergeometric Distribution:

The expected value (mean) and standard deviation for a hypergeometric distribution can be calculated as:

1. Mean (Expected Value): $E(X) = \frac{n \cdot K}{N}$ where:

   • $n = 20$ (number of tickets bought)
   • $K = 50$ (number of winning tickets)
   • $N = 500$ (total number of tickets)

   $E(X) = \frac{20 \cdot 50}{500} = 2$

2. Standard Deviation: $\sigma = \sqrt{\frac{n \cdot K \cdot (N - K) \cdot (N - n)}{N^2 \cdot (N - 1)}}$ where:

   • $n = 20$
   • $K = 50$
   • $N = 500$

   Plugging in the values: $\sigma = \sqrt{\frac{20 \cdot 50 \cdot (500 - 50) \cdot (500 - 20)}{500^2 \cdot (500 - 1)}}$ Simplifying: $\sigma = \sqrt{\frac{20 \cdot 50 \cdot 450 \cdot 480}{500^2 \cdot 499}} = \sqrt{\frac{2160000000}{124750000}} \approx 1.732$

Final Answer:

• The average number of winning tickets: 2
• The standard deviation: 1.732

Therefore, the correct answer is: 2 and 1.732

Would you like further clarification or details on the steps? Here are some related questions to deepen your understanding:

1. What is the difference between hypergeometric and binomial distributions?
2. How would the calculation change if the tickets were replaced (i.e., binomial distribution)?
3. How does the population size (N) affect the standard deviation in this context?
4. What would happen if the number of tickets bought (n) increased to 100?
5. Can you explain the formula for the standard deviation in the hypergeometric distribution in more detail?

Tip: For large population sizes and small sample sizes, the hypergeometric distribution can often be approximated by a binomial distribution, which is simpler to calculate.