One prominent physician claims that 70% of those with lung cancer are heavy smokers. If his assertion is correct, find the probability that out of 10 such patients recently admitted to a hospital

Question

(a) more than half are heavy smokers

(b) exactly 4 are heavy smokers

(c) less than 2 are non- smokers

Answer 1

use binomial formula, where probability x successes in n tries where each try has probability p:

P(x successes) = C(n,x) * p^x * (1-p)^(n-x)

here n = 10, p = 0.7

a) more than have is P(x = 6) + P(x=7) + P(x = 8) + P(x = 9) + P(x = 10)

b) exactly 4 are heavy-smokes is P(x = 4)

c) probability non-smoker is 1 - 0.7 = 0.3

P(x < 2) = P(x = 0) + P(x = 1), and P = 0.3