A new surgery is successful of the time. If the results of such surgeries are randomly sampled, what is the probability that more than of them are successful

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

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

go and ask your mom to you