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Suppose we have a test for a certain disease. If a person has the disease,the test has 95% chance to be positive. If a person is healthy, the test has 5% chanceto give a false positive result. We assume that a person has 10% chance to have thedisease. Given that the test for a person is positive, what is the probability that theperson has the disease?

Respuesta :

ridxee

Answer:

P ( disease | positive ) = 0.68

Step-by-step explanation:

This is a classic question of Bayes' Theorem, which calculates probability in a scenario where one event has already occured.

In this case, we have the following data:

P ( disease ) = 0.10

P (no disease) = 0.90

P ( positive | disease ) = 0.95

P ( positive | no disease) = 0.05

Formula to use:

[tex]P ( disease\ |\ positive ) = \frac{P(disease)\ P(positive\ |\ disease)}{P(disease)\ P(positive\ |\ disease) + P(no\ disease)\ P(positive\ |\ no\ disease)}[/tex]We substitute the values from our data in this formula:

[tex]\frac{(0.10) \times (0.95)}{(0.10)\times(0.95) + (0.90)\times(0.05)} \\\\\frac{0.095}{0.095+0.045}\\\\\frac{0.095}{0.14}\\ \\\\\=0.68[/tex] (Rounded off to two decimal places)

Hence, the probability of a person testing positive once they have the disease is 68%.