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75% of the people in a small city have car insurance with the company Geico. This unusually high amount is due to the amount of advertising withing the city and social media ads. If 10 people are selected at random from this city, what is the probability that at least 8 of them have car insurance with Geico?

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Answer:

Explanation:

This is a binomial distribution problem, where each person either has car insurance with Geico or doesn't have.

Let's denote the probability of a person having car insurance with Geico as p = 0.75, and the probability of not having car insurance with Geico as q = 0.25.

The probability of getting exactly k people out of n people who have car insurance with Geico is given by the binomial distribution formula:

P(k) = (n choose k) * p^k * q^(n-k)

where (n choose k) is the number of ways to choose k items out of n items, which is calculated by the formula:

(n choose k) = n! / (k! * (n-k)!)

where n! denotes the factorial of n.

We need to find the probability of getting at least 8 people out of 10 who have car insurance with Geico, which is the same as the probability of getting 8, 9, or 10 people who have car insurance with Geico. So, we can calculate this probability by adding the probabilities of these three cases:

P(at least 8) = P(8) + P(9) + P(10)

P(8) = (10 choose 8) * 0.75^8 * 0.25^2 = 0.002

P(9) = (10 choose 9) * 0.75^9 * 0.25^1 = 0.054

P(10) = (10 choose 10) * 0.75^10 * 0.25^0 = 0.056

Therefore, the probability of at least 8 people out of 10 having car insurance with Geico is:

P(at least 8) = P(8) + P(9) + P(10) = 0.002 + 0.054 + 0.056 = 0.112

So, the probability that at least 8 people out of 10 have car insurance with Geico is 0.112 or approximately 11.2%.

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