Math Problem Statement

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?

Solution

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=0.9p = 0.9.
  • The probability that a vaccinated person does contract the disease is 1p=10.9=0.11 - p = 1 - 0.9 = 0.1.
  • Let nn be the number of vaccinated people exposed to influenza A.
  • We are asked to find the smallest nn 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 0.9n0.9^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(at least one person contracts the disease)=10.9nP(\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: 10.9n0.951 - 0.9^n \geq 0.95

Step 3: Solve for nn

0.9n0.050.9^n \leq 0.05 Taking the natural logarithm of both sides: ln(0.9n)ln(0.05)\ln(0.9^n) \leq \ln(0.05) nln(0.9)ln(0.05)n \ln(0.9) \leq \ln(0.05) Since ln(0.9)\ln(0.9) is negative, dividing both sides by ln(0.9)\ln(0.9) reverses the inequality: nln(0.05)ln(0.9)n \geq \frac{\ln(0.05)}{\ln(0.9)}

Step 4: Compute the value of nn

Now, we can calculate the value of nn:

nln(0.05)ln(0.9)n \geq \frac{\ln(0.05)}{\ln(0.9)}

Let me compute this.The smallest number of vaccinated people, nn, 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.

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Math Problem Analysis

Mathematical Concepts

Probability
Complementary Probability
Logarithms
Exponential Decay

Formulas

P(at least one person contracts the disease) = 1 - 0.9^n
0.9^n ≤ 0.05
n ≥ ln(0.05) / ln(0.9)

Theorems

Complementary Probability Theorem
Laws of Logarithms

Suitable Grade Level

Grades 10-12