Question

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?

Answer 1

The probability that at least one out of four randomly selected people is vaccinated is calculated using the complement of the probability that none are vaccinated, which is determined by the vaccination rate of 58%.

Step-by-step explanation:

To find the probability that at least one person has been vaccinated when four people are randomly selected from a population where 58% have been vaccinated, we can use the complement rule. The complement rule states that the probability of at least one event occurring is equal to 1 minus the probability of none of the events occurring.

First, we calculate the probability that none of the four people are vaccinated. Since the vaccination rate is 58%, the non-vaccination rate is 1 - 0.58 = 0.42. The probability that all four people are not vaccinated is (0.42)⁴.

Now, we calculate the complement probability:

P(at least one is vaccinated) = 1 - P(none are vaccinated)

P(at least one is vaccinated) = 1 - (0.42)⁴

Calculating the value of (0.42)⁴ and subtracting it from 1 will give us the probability that at least one person out of the four selected is vaccinated.

Calculate the probability for X = 0: 0.42⁴ ≈ 0.028

Subtract this from 1 to find the probability of at least one person being vaccinated: 1 - 0.028 ≈ 0.972.

Both methods lead to the same result: the probability that at least one of the four people has been vaccinated is approximately 0.972 or 97.2%.

Answer 2

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