In a large population, 58 % of the people have been vaccinated. If 4 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

0.97

Step-by-step explanation:

From the question:

If 58% = 58/100 of the people have been vaccinated, then;

1 - (58/100) = 42% = 42/100 of the people have not been vaccinated.

Now:

Probability, P( > 1 ), that at least one of the selected four has been vaccinated is given by;

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

Where;

P(0) = probability that all of the four have not been vaccinated.

P(0) = P(1) x P(2) x P(3) x P(4)

Where;

P(1) = Probability that the first out of the four has not been vaccinated

P(2) = Probability that the second out of the four has not been vaccinated

P(3) = Probability that the third out of the four has not been vaccinated

P(4) = Probability that the fourth out of the four has not been vaccinated

Remember that 42/100 of the population have not been vaccinated. Therefore,

P(1) = 42/100    

P(2) = 42/100

P(3) = 42/100

P(4) = 42/100

P(0) = (42/100) x (42/100) x (42/100) x (42/100)

P(0) = (42/100)⁴

P(0) = (0.42)⁴

P(0) = 0.03111696

Therefore, from equation (1);

P( > 1 ) = 1 - 0.03111696

P( > 1 ) = 0.96888304 ≅ 0.97

Therefore, the probability that AT LEAST ONE of them has been vaccinated is 0.97

the first one, all those points on the line just make a strait line.