Question

g ) If it is raining, a home security system detects an intruder with probability 0.70. If it is NOT raining, the probability becomes 0.92. The probability of rain on any 2 given day is 0.25. To test the system on a randomly chosen day, the system technician pretends to be an intruder. Given that the technician will NOT be detected, what is the probability that it is NOT raining

Answer 1

Answer:

0.4444 = 44.44% probability that it is NOT raining

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: Technician not detected.

Event B: Not raining.

Probability the technician is not detected:

0.3 of 0.25(raining).

0.08 of 0.75(not raining). So

$P(A) = 0.3*0.25 + 0.08*0.75 = 0.135$

Probability the technician is not detected and it is not raining:

0.08 of 0.75. So

$P(A \cap B) = 0.08*0.75 = 0.06$

Given that the technician will NOT be detected, what is the probability that it is NOT raining?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.135} = 0.4444$

0.4444 = 44.44% probability that it is NOT raining