A carrier of tuberculosis has a 10% chance of passing the disease on to anyone with whom he comes into close contact who has had no prior exposure. During a day, he meets 10 such individuals. Calculate the probability that less than 6 of these individuals will contract

Answer 1

Tuberculosis from the carrier.

Since each close contact has a 10% chance of contracting tuberculosis, this can be modeled as a binomial distribution with n=10 (number of trials) and p=0.1 (probability of success).

We want to calculate the probability that less than 6 individuals will contract tuberculosis from the carrier. This can be calculated by summing the probabilities of 0, 1, 2, 3, 4, and 5 individuals contracting tuberculosis:

P(X < 6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

Using the binomial probability formula:

P(X=k) = (n choose k) * p^k * (1-p)^(n-k)

where (n choose k) is the number of ways to choose k successes out of n trials:

P(X=0) = (10 choose 0) * 0.1^0 * 0.9^10
P(X=1) = (10 choose 1) * 0.1^1 * 0.9^9
P(X=2) = (10 choose 2) * 0.1^2 * 0.9^8
P(X=3) = (10 choose 3) * 0.1^3 * 0.9^7
P(X=4) = (10 choose 4) * 0.1^4 * 0.9^6
P(X=5) = (10 choose 5) * 0.1^5 * 0.9^5

Calculating each of these probabilities and summing the results:

P(X < 6) ≈ 0.9884

Therefore, the probability that less than 6 individuals will contract tuberculosis from the carrier is approximately 0.9884.