Question

In a large population, 52 % of the people have been vaccinated. If 3 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

Answer 1

Solution

Given that ,

p = 0.52

1 - p = 0.48

n = 3

x 1

Using binomial probability formula ,

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

P(X 1 ) = P(X=1)+P(X=2)+P(X=3)

= ((3! / 1! (3-1)!) * 0.52¹ * (0.48)^3-1 + ((3! / 2! (3-2)!) * 0.52² * (0.48)^3-2 + ((3! / 3! (3-3)!) * 0.52³ * (0.48)^3-3

=  ((3! / 1! (2)!) * 0.52¹ * (0.48)² + ((3! / 2! (1)!) * 0.52² * (0.48)¹ + ((3! / 3! (0)!) * 0.52³ * (0.48)⁰

=  0.3594 + 0.3894 + 0.1406

Probability = 0.8894