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 fi

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?

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Thiên Hương 4 years 2021-08-27T10:40:51+00:00 1 Answers 13 views 0

Answers ( )

    0
    2021-08-27T10:42:50+00:00

    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

    P(B|A) = \frac{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) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967

    0.967 = 96.7% probability the rock sample actually contains raritanium

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