Question

A surgical technique is performed on seven patients. You are told there is a 70% chance of success. Use a table to find the probability the surgery is successful for exactly five patients.

Answer 1

Answer:

31.77% probability the surgery is successful for exactly five patients.

Step-by-step explanation:

For each patient, there are only two possible outcomes. Either the surgery is successful, or it is not. The probability of the surgery being successful for a patient is independent of other patients. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A surgical technique is performed on seven patients.

This means that $n = 7$

You are told there is a 70% chance of success.

This means that $p = 0.7$

Find the probability the surgery is successful for exactly five patients.

This is P(X = 5).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{7,5}.(0.7)^{5}.(0.3)^{2} = 0.3177$

31.77% probability the surgery is successful for exactly five patients.