Question

A surgical technique is performed on eight patients. You are told there is an 80% chance of success. Find the probability that the surgery is successful for fewer than two patients. Please show your work.

Answer 1

0.0344%

Explanation:

We want to calculate the probability of having fewer than two successful surgeries (0 or 1 successful surgery) out of eight patients, given that each surgery has an 80% chance of success (p = 0.8) and a 20% chance of failure (1 - p = 0.2).

P(X = 0) = C(8, 0) * 0.8^0 * 0.2^8

P(X = 1) = C(8, 1) * 0.8^1 * 0.2^7C(8, 0) = 1

C(8, 1) = 8P(X = 0) = 1 * 1 * (0.2)^8 = 0.000016384

P(X = 1) = 8 * 0.8 * (0.2)^7 = 0.00032768

Now, add these probabilities together to get the final result:

P(X < 2) = P(X = 0) + P(X = 1) = 0.000016384 + 0.00032768 ≈ 0.000344064

The correct probability that the surgery is successful for fewer than two patients is approximately 0.0344% or 0.000344064. This means that the chance of having fewer than two successful surgeries out of eight is very low.