Question

When a new highway is formed, the smoothness of the surface must be verified. A contractor is making a bid for this job and has a truck with the relevant detector. This detector must meet minimum standards for detecting irregularities in the road surface. If there is a roadway irregularity, there is a probability of 0.99 the detector will detect it. If there is no irregularity, there is a 0.06 probability detector will identify it as irregular (a false positive). It is known through experience that 3 miles out of 100 miles actually contain irregularities.
a. What is the probability the detector will identify a random mile of roadway as irregular? Give your answer to four decimal places.
b. Given a randomly selected miles has been identified as irregular by the detector, what is the probability it actually is irregular? Give your answer to four decimal places.

Answer 1

Answer:

a) 0.0879 = 8.79% probability the detector will identify a random mile of roadway as irregular.

b) 0.3379 = 33.79% probability it actually is irregular

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.

a. What is the probability the detector will identify a random mile of roadway as irregular?

99% of 3%(it is irregular).

6% of 97%(false positive). So

$p = 0.99*0.03 + 0.06*0.97 = 0.0879$

0.0879 = 8.79% probability the detector will identify a random mile of roadway as irregular.

b. Given a randomly selected miles has been identified as irregular by the detector, what is the probability it actually is irregular? Give your answer to four decimal places.

Conditional probability:

Event A: Identified as irregular

Event B: It is irregular.

0.0879 = 8.79% probability the detector will identify a random mile of roadway as irregular, which means that $P(A) = 0.0879$

99% of 3% arre irregulars identified as, which means that $P(A \cap B) = 0.03*0.99 = 0.0297$

The desired probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0297}{0.0879} = 0.3379$

0.3379 = 33.79% probability it actually is irregular