Question

Po is an oncologist with seven patients in their care. the probability that a patient will survive five years after being diagnosed with stage three breast cancer is 0.82. what is the probability that four of the patients are still alive after five years?

Answer 1

Answer:

0.0923 = 9.23% probability that four of the patients are still alive after five years.

Step-by-step explanation:

For each patient, there are only two possible outcomes. Either they are still alive after five years, or they are not. The probability of a patient being alive is independent of any other patient, which means that the binomial probability distribution is used 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.

Po is an oncologist with seven patients in their care.

This means that $n = 7$

The probability that a patient will survive five years after being diagnosed with stage three breast cancer is 0.82.

This means that $p = 0.82$

What is the probability that four of the patients are still alive after five years?

This is P(X = 4). So

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

$P(X = 4) = C_{7,4}.(0.82)^{4}.(0.18)^{3} = 0.0923$

0.0923 = 9.23% probability that four of the patients are still alive after five years.