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Vikentia [17]
3 years ago
5

In a large population, 69% of the people have been vaccinated. If 5 people are randomly selected, what is the probability that a

t least one of them has been vaccinated?
Mathematics
1 answer:
Darya [45]3 years ago
3 0

Answer: 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

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Otherwise, if x - 4 < 0, then |x - 4| = -(x - 4), so

G(x) = F(x)   ⇒   3x + 2 = -(x - 4) + 2

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⇒   4x = 4

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However,

• when x = -2, we have

G(-2) = 3(-2) + 2 = -4

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so only x = 1 is a solution to G(x) = F(x).

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2 years ago
Please Help!! If You Answer First (And It's Good) Then I'll Mark You Brainly! &lt;3
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3 years ago
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2 years ago
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Answer:

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