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.)

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

add an image, and i will answer

step-by-step explanation:

stops on an even number.