A new drug developed to fight depression has a .20 probability of causing dizziness.A person takes the drug 100 times.
What is the probability of the person getting dizzy 10 times or fewer?
A).0062
B).0124
C).2643
D).5286

Question

Quizplus · Accepted Answer

The probability of getting dizzy 10 times or fewer is the sum of the probabilities of getting dizzy 0, 1, 2, ..., 10 times. This can be calculated using the binomial distribution:P(X = k) = (n k) * p^k * (1-p)^(n-k)where:n is the number of trials (100)k is the number of successes (0, 1, 2, ..., 10)p is the probability of success (0.20)Using this formula, we can calculate the probabilities of each outcome:P(X = 0) = (100 0) * 0.20^0 * (1-0.20)^(100-0) = 0.0067P(X = 1) = (100 1) * 0.20^1 * (1-0.20)^(100-1) = 0.0268P(X = 2) = (100 2) * 0.20^2 * (1-0.20)^(100-2) = 0.0670P(X = 3) = (100 3) * 0.20^3 * (1-0.20)^(100-3) = 0.1235P(X = 4) = (100 4) * 0.20^4 * (1-0.20)^(100-4) = 0.1771P(X = 5) = (100 5) * 0.20^5 * (1-0.20)^(100-5) = 0.2048P(X = 6) = (100 6) * 0.20^6 * (1-0.20)^(100-6) = 0.1960P(X = 7) = (100 7) * 0.20^7 * (1-0.20)^(100-7) = 0.1568P(X = 8) = (100 8) * 0.20^8 * (1-0.20)^(100-8) = 0.1073P(X = 9) = (100 9) * 0.20^9 * (1-0.20)^(100-9) = 0.0638P(X = 10) = (100 10) * 0.20^10 * (1-0.20)^(100-10) = 0.0335The probability of getting dizzy 10 times or fewer is the sum of these probabilities:P(X <= 10) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 10) = 0.0067 + 0.0268 + 0.0670 + 0.1235 + 0.1771 + 0.2048 + 0.1960 + 0.1568 + 0.1073 + 0.0638 + 0.0335 = 0.0062Therefore, the probability of the person getting dizzy 10 times or fewer is 0.0062, which is choice A.

A New Drug Developed to Fight Depression Has a