Question

Suppose that babies are born at an average rate of 5 per day at a certain hospital.

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a.What is the probability that on any given day, exactly one baby is born?

b. The probability that a patient recovers from a rare blood disease is 0.6. If 7 people are known to have contracted this disease, what is the probability that at least 5 survive?

Answer 1

The probability of exactly one baby being born on any given day is 9.15%.
The probability of at least 5 people surviving from a rare blood disease is 49.05%.

Step-by-step explanation:

a. Probability of Exactly One Baby Born

To find the probability of exactly one baby being born on any given day, we can use the Poisson distribution. The average rate is 5 babies per day, so the parameter λ (lambda) is also 5. The probability of exactly one baby being born is given by the formula:
P(X = 1) = (e^(-λ) * λ^1) / 1! = (e^(-5) * 5^1) / 1! = (1.83 * 5) / 1 = 9.15%

b. Probability of At Least 5 Surviving from Rare Blood Disease

To find the probability of at least 5 people surviving from a rare blood disease, we can use the binomial distribution. The probability of a patient surviving is 0.6, and we have 7 people with the disease. We want to find P(X ≥ 5). Using the binomial probability formula:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) = (7 choose 5) * (0.6^5) * (0.4^2) + (7 choose 6) * (0.6^6) * (0.4^1) + (7 choose 7) * (0.6^7) * (0.4^0) = 0.29376 + 0.15606 + 0.04096 = 49.05%