Question

In a large population, 69 % 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

Answer:

$P( X \geq 1)=1-P(X$

And the probability is:

$P(X=0) = (3C0) (0.69)^0 (1-0.69)^{3-0} = 0.0298$

And replacing we got:

$P( X \geq 1)=1-P(X$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=3, p=0.69)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Solution to the problem

For this case we want the following probability:

$P( X \geq 1)$

And we can calculate thi with the complement rule:

$P( X \geq 1)=1-P(X$

And the probability is:

$P(X=0) = (3C0) (0.69)^0 (1-0.69)^{3-0} = 0.0298$

And replacing we got:

$P( X \geq 1)=1-P(X$