Question

there are two mutually ezclusive reasonsfor visiting the emergency room of rhe local hospital: it is either an emergency or its not an emergency. If the probability that a patient visiting the emergency. eoom has an emergency os 0.88, what is the probability that the next patient hs an emergency and the patient after the next ome does not have an emergency? assume that the patients arrive at the emergency room independently. (round to three decimal places

Answer 1

If the probability of emergency is X . And the probability of not an emergency is (1-X) .

We want to find ,

the probability that the next patient has an emergency and the patient after the next one does not have an emergency is =X*(1-X)

X=0.88

1-X=0.12

Answer is =X*(1-X)=0.88*0.12=0.106

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Answer 2

Let's denote the event "visiting the emergency room" as E and the event "having an emergency" as A. We are given that the probability of a patient visiting the emergency room having an emergency is 0.88, so P(A|E) = 0.88.

We want to find the probability that the next patient has an emergency (A) and the patient after the next one does not have an emergency (~A).

Since the patients arrive independently, we can use the multiplication rule for independent events:

P(A and ~A) = P(A) * P(~A)

We know that P(A|E) = 0.88, so P(A) represents the probability of having an emergency, regardless of whether the patient visited the emergency room or not.

P(~A) represents the probability of not having an emergency. Since there are only two mutually exclusive possibilities (emergency or not emergency), P(~A) = 1 - P(A).

Let's calculate the probability:

P(A and ~A) = P(A) * P(~A) = P(A) * (1 - P(A)) = 0.88 * (1 - 0.88) = 0.88 * 0.12 = 0.1056

Therefore, the probability that the next patient has an emergency and the patient after the next one does not have an emergency is approximately 0.106 (rounded to three decimal places).