From experience the manager of KramerSolutionbinomial distri

From experience, the manager of Kramer

binomial distribution with p = 0.40 and n = 10
P(x) = nCx * (p^x) * ((1-p)^(n-x))

probability that among ten people who are browsing, at least three will make a purchase
P(X>=3)
= 1 - (P(0) + P(1) + P(2))
= 1 - ((10C0 * 0.40^0 * 0.60^10) + (10C1 * 0.40^1 * 0.60^9) + (10C2 * 0.40^2 * 0.60^8))
= 1 - ((1 * 0.40^0 * 0.60^10) + (10 * 0.40^1 * 0.60^9) + (45 * 0.40^2 * 0.60^8))
= 0.8327