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)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D)
In which
P(B|A) is the probability of event B happening, given that A happened.
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](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.98%2A0.13%20%2B%200.005%2A0.87%20%3D%200.13175)
Positive when there is raritanium:
98% of 13%
![P(A) = 0.98*0.13 = 0.1274](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.98%2A0.13%20%3D%200.1274)
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](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.1274%7D%7B0.13175%7D%20%3D%200.967)
0.967 = 96.7% probability the rock sample actually contains raritanium