The efficacy of the influenza A vaccine is about 90%; that is, 90% of those receiving the influenza A vaccine will not contract the disease when exposed. Assume that each person’s response to influenza A is independent of another person’s response. How many vaccinated people must be exposed to the influenza A so that the probability of at least one person contracting the disease is at least 0.95?


This problem can be approached using the concept of complementary probability and the binomial distribution.

### Step 1: Define the variables and probabilities
- The efficacy of the influenza A vaccine is 90%, so the probability that a vaccinated person **does not** contract the disease is \( p = 0.9 \).
- The probability that a vaccinated person **does** contract the disease is \( 1 - p = 1 - 0.9 = 0.1 \).
- Let \( n \) be the number of vaccinated people exposed to influenza A.
- We are asked to find the smallest \( n \) such that the probability of **at least one** person contracting the disease is **at least 0.95**.

### Step 2: Complementary probability approach
The probability that **no one** contracts the disease is \( 0.9^n \) because each individual has a 90% chance of not contracting the disease, and the exposures are independent.

Thus, the probability that **at least one person** contracts the disease is:
\[
P(\text{at least one person contracts the disease}) = 1 - 0.9^n
\]
We want this probability to be at least 0.95, so we set up the inequality:
\[
1 - 0.9^n \geq 0.95
\]

### Step 3: Solve for \( n \)
\[
0.9^n \leq 0.05
\]
Taking the natural logarithm of both sides:
\[
\ln(0.9^n) \leq \ln(0.05)
\]
\[
n \ln(0.9) \leq \ln(0.05)
\]
Since \( \ln(0.9) \) is negative, dividing both sides by \( \ln(0.9) \) reverses the inequality:
\[
n \geq \frac{\ln(0.05)}{\ln(0.9)}
\]

### Step 4: Compute the value of \( n \)
Now, we can calculate the value of \( n \):

\[
n \geq \frac{\ln(0.05)}{\ln(0.9)}
\]

Let me compute this.The smallest number of vaccinated people, \( n \), such that the probability of at least one person contracting the disease is at least 0.95, is approximately 29 people.

Would you like more details or have any other questions?

Here are 5 related questions:
1. What would the number of people be if the efficacy were 80% instead of 90%?
2. How does the probability of at least one person contracting the disease change if the exposure group size increases?
3. Can this approach be generalized for other diseases with different vaccine efficacies?
4. How would the result change if we increased the threshold probability from 0.95 to 0.99?
5. What are the assumptions made when modeling this problem using independent events?

**Tip:** When using complementary probability, calculating the chance of "none" happening often simplifies solving for "at least one" occurrences.

The efficacy of the influenza A vaccine is about 90%; that is, 90% of those receiving the influenza A vaccine will not contract the disease when exposed. Assume that each person’s response to influenza A is independent of another person’s response. How many vaccinated people must be exposed to the influenza A so that the probability of at least one person contracting the disease is at least 0.95?

Find out how to calculate the minimum number of vaccinated people needed for at least one person to contract the disease with a 0.95 probability threshold. Using complementary probability and logarithms, this problem explores key concepts in probability and exponential decay. Suitable for Grades 10-12.

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