True or false? on average, 20% of the emergency room patients at greenwood general hospital lack health insurance. in a random sample of 4 patients, the probability that only one patient will be uninsured is 0.4096.

Question

Jamya

Biology

11 December, 17:51

True or false? on average, 20% of the emergency room patients at greenwood general hospital lack health insurance. in a random sample of 4 patients, the probability that only one patient will be uninsured is 0.4096.

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Answers (1)

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Celia Espinoza · Answer 1

To solve this, we use the binomial probability equation.

P = [n! / (n - r) ! r!] p^r * q^ (n - r)

where n is the total number of sample patients = 4, r is the number of uninsured patients = 1, p is probability of being uninsured = 0.20, q is 1 - p = 0.80

Calculating for probability P:

P = [4! / (4 - 1) ! 1!] 0.20^1 * 0.80^ (4 - 1)

P = 0.4096

Answer:

True