According to a recent survery, in 2024 about 20% of the people in McAllen, did not have health insurance. If we select 8 people from Mcallen at random, what is the probability that 1,2, or 3,of them do not have health insurance?
Question
According to a recent survery, in 2024 about 20% of the people in McAllen, did not have health insurance. If we select 8 people from Mcallen at random, what is the probability that 1,2, or 3,of them do not have health insurance?
Solution
This problem can be solved using the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case, the probability of not having health insurance (p) is 0.20 (or 20%), the number of trials (n) is 8 (the number of people selected), and the number of successes (k) is 1, 2, or 3 (the number of people without health insurance).
We need to calculate the probability for k=1, k=2, and k=3, and then add these probabilities together to get the total probability.
-
For k=1: P(X=1) = C(8, 1) * (0.20^1) * ((1-0.20)^(8-1))
-
For k=2: P(X=2) = C(8, 2) * (0.20^2) * ((1-0.20)^(8-2))
-
For k=3: P(X=3) = C(8, 3) * (0.20^3) * ((1-0.20)^(8-3))
Finally, add these three probabilities together to get the total probability:
P = P(X=1) + P(X=2) + P(X=3)
This will give you the probability that 1, 2, or 3 people out of 8 selected at random from McAllen do not have health insurance.
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