Question

(10 pts) In a large town, one person in 80, on average, has blood type X. If 200 blood donors are taken at random, find an approximation to the probability that they include at least five persons having blood type X.

How many donors must be taken at random in order that the probability of including at least one donor of type X shall be 0.9 or more? ​

Answer 1

Solution:

Poisson approximation X~ Po(λ) where λ = 200* 1/8 = 2.5

P(X ≥ 5) = 1- P(X ≤ 4)

= 1- (P(X = 0) + P(X = 1) + P(X = 3) + P(X = 4))

= 1- (e^-2.5 (2.5⁰/0!)+ e^-2.5 (2.5¹/1!) + e^-2.5 (2.5²/2!)+ e^-2.5 (2.5³/3!) +e^-2.5 (2.5⁴/4!))

= 0.1088220

The donors must be taken at random in order that the probability of including at least one donor of type X shall be 0.9 or more is

n donors => λ = n 1/80

P(X ≥ 1) = 1- P(X = 0) = 1- e^(n/80)((n/80)⁰/0!) = 1- e^(n/80)

We want P(X ≥ 1) ≥ 0.9

1- e^(n/80) ≥ 0.9

=> n ≥ 184.2

So we need 185 donors