This problem can be approached using probability principles, particularly binomial probability and the complement rule.

Part 1: Probability that at least one person is infected

Let’s define the situation:

• The probability that a person is not infected is $P(\text{not infected}) = \frac{99}{100} = 0.99$.
• The probability that a person is infected is $P(\text{infected}) = \frac{1}{100} = 0.01$.

To find the probability that at least one person in the group of 40 is infected, we use the complement rule. First, we calculate the probability that no one is infected, and then subtract that from 1.

• The probability that all 40 people are not infected is: $P(\text{no one infected}) = 0.99^{40}$
• The probability that at least one person is infected is: $P(\text{at least one infected}) = 1 - P(\text{no one infected}) = 1 - 0.99^{40}$

Let’s calculate $0.99^{40}$.

$0.99^{40} \approx 0.6702$

So, the probability that at least one person is infected is:

$P(\text{at least one infected}) \approx 1 - 0.6702 = 0.3298$

Thus, the probability that at least one person out of 40 is infected is approximately $0.3298$, or 32.98%.

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Part 2: Probability that at least 2 people are infected

To find the probability that at least two people are infected, we first calculate the probabilities that 0 or 1 person is infected, and subtract that from 1.

1. Probability that no one is infected: We already know this is $P(\text{no one infected}) = 0.99^{40} \approx 0.6702$.

2. Probability that exactly 1 person is infected:

   This follows a binomial probability formula: $P(\text{exactly 1 infected}) = \binom{40}{1} \cdot (0.01)^1 \cdot (0.99)^{39}$ $P(\text{exactly 1 infected}) = 40 \cdot (0.01) \cdot (0.99)^{39}$

   Now, we calculate $(0.99)^{39}$.

$0.99^{39} \approx 0.6760$

Thus, the probability of exactly one person being infected is:

$P(\text{exactly 1 infected}) = 40 \cdot 0.01 \cdot 0.6760 \approx 0.2704$

3. Probability that at least two people are infected: $P(\text{at least two infected}) = 1 - P(\text{no one infected}) - P(\text{exactly 1 infected})$ $P(\text{at least two infected}) = 1 - 0.6702 - 0.2704 \approx 1 - 0.9406 = 0.0594$

Thus, the probability that at least two people out of 40 are infected is approximately $0.0594$, or 5.94%.

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Summary:

• The probability that at least one of the 40 people is infected is approximately 32.98%.
• The probability that at least two of the 40 people are infected is approximately 5.94%.

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Would you like more details or have any questions on this?

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Here are 5 questions that build on this:

1. What is the probability that exactly 3 people are infected out of 40?
2. How would the probability change if the infection rate was 2% instead of 1%?
3. How does the binomial distribution apply to this scenario?
4. Can you calculate the expected number of infected people in a group of 40?
5. What would be the variance of the number of infected people in a group of 40?

Tip: When dealing with probabilities involving "at least" scenarios, the complement rule can simplify the problem by focusing on the opposite event.