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

Answer:

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

Answer 2

The probability will be "0.969".

According to the question,

Population,

• P = 58%

or,

          = 0.58

Number of people,

• n = 4

In a Binomial distribution,

→ $P(X=x )= n_C_x P^x q^{n-x}$

where,

→ $q = 1-P$

     $= 0.42$

∴ X = at least one

hence,

→ $P(x \geq 1) = 1-P(X<1)$

                  $= 1-P(x=0)$

                  $=1-4_C_0 (0.58)^0 (0.42)^4$

                  $= 1-0.03112$

                  $= 0.969$

Thus the answer above is right.

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