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?

Question

contestada

Respuesta :

joaobezerra joaobezerra · Answer 1 · 2021-02-15T19:39:46+01:00

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

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

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

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

[tex]P(A \cap B)[/tex] 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

[tex]P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175[/tex]

Positive when there is raritanium:

98% of 13%

[tex]P(A) = 0.98*0.13 = 0.1274[/tex]

What is the probability the rock sample actually contains raritanium?

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967[/tex]

0.967 = 96.7% probability the rock sample actually contains raritanium