Question

The probability that the patient recovers from a rare deadly disease is 40% and failure to recover is 60%.If 15 people contracted this disease what is the probability that a. 5 will survive? b. 8 will survive? c.12 will survive?(complete solution)​

Answer 1

Answer:

a) 0.1859 = 18.59% probability that 5 will survive.

b) 0.1181 = 11.81% probability that 8 will survive.

c) 0.0016 = 0.16% probability that 12 will survive.

Step-by-step explanation:

For each patient, there are only two possible outcomes. Either they survive, or they do not. The probability of a patient surviving is independent of any other patient. This 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.

The probability that the patient recovers from a rare deadly disease is 40%

This means that $p = 0.4$

15 individuals

This means that $n = 15$

a. 5 will survive

This is P(X = 5). So

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

$P(X = 5) = C_{15,5}.(0.4)^{5}.(0.6)^{10} = 0.1859$

0.1859 = 18.59% probability that 5 will survive.

b. 8 will survive

This is P(X = 8). So

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

$P(X = 8) = C_{15,8}.(0.4)^{8}.(0.6)^{7} = 0.1181$

0.1181 = 11.81% probability that 8 will survive.

c.12 will survive

This is P(X = 12). So

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

$P(X = 12) = C_{15,12}.(0.4)^{12}.(0.6)^{3} = 0.0016$

0.0016 = 0.16% probability that 12 will survive.