Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i

Question

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Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i

s 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.)

Mathematics

1 answer:

Zina [12.3K] · Answer 1 · 2026-03-02T03:01:53+03:00

Answer:

The likelihood that the rock contains raritanium is 0.9670

Step-by-step explanation:

Define the following events

Z: the sample contains raritanium

Y: the detector's reading is positive

Z': the sample does not have raritanium

Y': the detector's reading is negative

P(z) = 0.13

P(y'/z) = 0.02

P(y/z')= 0.005

We aim to determine p(z/y)

= p(z/y) = p(z ∩ y)/p(y)

= P(z) - p(z∩y)/0.13 = 0.02

Keep in mind p(z) = 0.13

Applying cross multiplication gives us

0.13 - p(z ∩ y) = 0.02 x 0.13

0.13 - p(z ∩ y) = 0.0026

-p(z ∩ y) = 0.0026 - 0.13

Therefore,

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

Cross multiplying gives

P(y) - 0.1274 = 0.005x0.87

P(y) = 0.1274+0.00435

P(y) = 0.13175

We assess 0.1274/0.13175

= 0.9670

Hence, we conclude that the probability of the rock containing raritanium is 0.9670