Complete question is;
A new surgery is successful 85% of the time. If the results of 7 such surgeries are randomly sampled, what is the probability that more than 4 of them are successful?
Carry your intermediate computations to at least four decimal places, and round your answer to at least two decimal places
Answer:
P(X > 4) = 0.93
Step-by-step explanation:
This is a binomial probability distribution problem with the formula;
P(X = x) = C(n, x) × p^(x) × (1 - p)^(n - x)
We are told that a new surgery is successful 85% of the time. Thus;
p = 85% = 0.85
n = 7
probability that more than 4 of them are successful would be;
P(X > 4) = P(5) + P(6) + P(7)
P(5) = C(7,5) × 0.85^(5) × (1 - 0.85)^(7 - 5)
P(5) = 0.2097
P(6) = C(7,6) × 0.85^(6) × (1 - 0.85)^(7 - 6)
P(6) = 0.3960
P(7) = C(7,7) × 0.85^(7) × (1 - 0.85)^(7 - 7)
P(5) = 0.3206
P(X > 4) = 0.2097 + 0.3960 + 0.3206
P(X > 4) = 0.9263 ≈ 0.93
