Question

A new surgery is successful80% of the time. If the results of7 such surgeries are randomly sampled, what is the probability that more than 5 of them are successful? Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Answer 1

This is a case of binomial distribution.

Probability of a successful surgery = 0.8

Total number of surgeries = 7

So, the probability that there are more than 5 successful surgeries out of randomly 7 selected surgeries =

Let X be the random variable denoting the number of successful surgeries.

So, P(X > 5) = P(X = 6) + P(X = 7)

= ⁷C₆ * 0.8⁶ * 0.2¹+ ⁷C₇ * 0.8⁷ * 0.2⁰

= 0.367 + 0.2097

= 0.5767

So, the probability that more than 5 of them are successful is 0.5767.

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