Question

If the probability of developing drug-induced lupus for a specific drug is .025 and 25 individuals under a particular doctor's care are currently taking the drug, what is the probability that exactly 7 individuals will develop drug-induced lupus from the drug?

Answer 1

Solution

Given that ,

p = 0.025

1 - p = 1 - 0.025 = 0.975

n = 25

x = 7

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X = 7) = ((25! / 7! (25 - 7)!) * 0.025³ * (0.975)^{25 - 7}

= ((25! / 7! (18)!) * 0.025³ * (0.975)¹⁸

= 0.00000186

Probability = 0.00000186