Question

In a large population, 68 % 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?

Give your answer as a decimal (to at least 3 places) or fraction.   

Answer 1

We use binomial distribution

X: The people have been vaccinated

p = probability of success for vaccination = 0.68 ( 68% in converted into decimal )

n = random sample size = 3

At least one means greater than or equal to 1

We have to find

$P(X \geq 1)$

We use complement rule

P(X > 1) = 1- P(X = 0)

$P(X=x) = \binom{n}{x} *p^{x}*(1-p)^{(n-x)}$

$\binom{n}{x} = \frac{n!}{x!*(n-x)!}$

$P(X=0) = \binom{3}{0} *0.68^{0}*(1-0.68)^{(3-0)}$

$\binom{3}{0} = \frac{3!}{0!*(3-0)!}$

0! is always 1 because we can arrange 0 by 1 way

$\binom{3}{0} = \frac{3!}{1*(3!)}$

$\binom{3}{0} = \frac{3*2*1}{1*(3*2*1)}$

$\binom{3}{0} = 1$

$0.68^{0}=1$

Now we use formula

P(X > 1) = 1- P(X = 0)

$P(X\geq1) =1- 0.032768 = 0.967232$

$P(X\geq1)= {\color{Purple}\mathbf{ 0.967232}}$

The probability that at least one of them has been vaccinated = 0.967232

Round answer to 3 decimal place

The probability that at least one of them has been vaccinated = 0.967

Final answer:-

The probability that at least one of them has been vaccinated = 0.967