Final answer:
The probability that exactly ten of the exposed people will contract the disease is 0.120
Step-by-step explanation:
This is a binomial experiment since all three characteristics are met. Each person is a trial. Since we sample 200 people, the number of trials is n = 200. For each person, there are two possible outcomes: will develop the disease or not. Since we are measuring the number of people who will develop the disease, a person who will develop the disease is defined as a success and a person who will not develop the disease is defined as a failure. The risk of developing the disease is 1.28 percent. Therefore the probability of a success, p = 1.28 percent, .0128, and the probability of a failure, q = 1 − p = 1.0128 = .9872. Both P and q are fixed over these 200 trials since we don't change anything and we don't have any reason to think the underlying disease is changing.
The formula for the probability of exactly x successes in n trials is:
P(X = x) = C(n, x) * p^x * q^(n-x)
Where C(n, x) is the combinations formula for choosing x items from a set of n. In this case, we want to find the probability that exactly ten of the exposed people will contract the disease:
P(X = 10) = C(15, 10) * 0.0128^10 * 0.9872^5
Calculating this gives us:
P(X = 10) = 3003 * 0.0128^10 * 0.9872^5 = 0.119 option B (0.120)