In a large population, 69% of the people have been vaccinated. If 5 people are randomly selected, what is the probability that at least one of them has been vaccinated?

P(x ≥ 1) = 0.997

Step-by-step explanation:

We would assume a binomial distribution for the the people that were sampled. The formula is expressed as

P(x = r) = nCr × p^r × q^(n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - r) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 69% = 69/100 = 0.69

q = 1 - p = 1 - 0.69

q = 0.31

n = 5

Therefore,

P(x ≥ 1) = 1 - P(x = 0)

P(x = 0) = 5C0 × 0.69^0 × 0.31^(5 - 0) = 0.003

P(x ≥ 1) = 1 - 0.003 = 0.997

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

⭐⭐

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step-by-step explanation: