Question

Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector is used to find rock samples that may contain the valuable mineral, but it is not perfect: When applied to a rock sample, there is a 2% chance of a false negative; that is, of a negative reading given raritanium is actually present. Moreover there is a 0.5% chance of a false positive; that is, of a positive reading when in fact no raritanium is there. Assume that 13% of all rock samples contain raritanium. The detector is applied to a sample and returns a positive reading.

Required:
What is the probability the rock sample actually contains raritanium?

Answer 1

0.967 = 96.7% probability the rock sample actually contains raritanium

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

`P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175`

Positive when there is raritanium:

98% of 13%

`P(A) = 0.98*0.13 = 0.1274`

What is the probability the rock sample actually contains raritanium?

`P(B|A) = (P(A \cap B))/(P(A)) = (0.1274)/(0.13175) = 0.967`

0.967 = 96.7% probability the rock sample actually contains raritanium