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. What is the probability the rock sample actually contains raritanium? (Give the answer rounded to at least four decimal places.)

Answer 1

Answer:

the probability the rock has raritanium is 0.9670

Step-by-step explanation:

These events have been defined as

Z: sample has raritanium

Y: the reading from detector is positive

Z': sample has no raritanium

Y': Reading from detector is negative

P(z) = 0.13

P(y'/z) = 0.02

P(y/z')= 0.005

We need to find p(z/y)

= p(z/y) = p(z ∩ y)/p(y)

= P(z) - p(z∩y)/0.13 = 0.02

Remember the value of p(z) = 0.13

So when we cross multiply we get

0.13 - p(z ∩ y) = 0.02 x 0.13

0.13 - p(z ∩ y) = 0.0026

-p(z ∩ y) = 0.0026 - 0.13

Such that

p(z ∩ y) = 0.1274

P(y/z')= 0.005

P(z∩y') = 0.005

Since p(z) = 0.13

P(z') = 1-0.13

P(y) - p(z∩y)/0.87 = 0.005

We cross multiply

P(y) - 0.1274 = 0.005x0.87

P(y) = 0.1274+0.00435

P(y) = 0.13175

We have 0.1274/0.13175

= 0.9670

So we conclude that probability rock has raritanium is 0.9670