A scooter factory runs two assembly lines, A and B. 97.4% of line A’s scooters pass instruction, while only 94.9% of line B’s scooters pass inspection. 65% of the factory’s scooters come off assembly line A.
a) What is the probability that a randomly selected scooter did not pass inspection?
b) What is the probability that if a randomly selected scooter passes inspection, it came from
assembly line B?
To find the probabilities, we need to use conditional probability.
a) To find the probability that a randomly selected scooter did not pass inspection, we can use the complement rule. The complement of passing inspection is failing inspection.
Let's denote the event "scooter passing inspection" as P and the event "scooter failing inspection" as ¬P.
We are given that 97.4% of line A's scooters pass inspection, so the probability of passing inspection for a scooter from line A, denoted as P(A), is 0.974. Therefore, the probability of a scooter from line A failing inspection, denoted as ¬P(A), is 1 - 0.974 = 0.026.
Similarly, for line B, we are given that 94.9% of line B's scooters pass inspection, so the probability of passing inspection for a scooter from line B, denoted as P(B), is 0.949. Therefore, the probability of a scooter from line B failing inspection, denoted as ¬P(B), is 1 - 0.949 = 0.051.
However, we are also given that 65% of the factory's scooters come off assembly line A, which implies that 35% of the factory's scooters come off assembly line B. So, the probability of selecting a scooter from line A, denoted as P(A), is 0.65, and the probability of selecting a scooter from line B, denoted as P(B), is 0.35.
Using the law of total probability, we can find the probability of a scooter failing inspection, denoted as ¬P, by summing the probabilities of a scooter failing inspection from line A and line B:
¬P = P(A) * ¬P(A) + P(B) * ¬P(B)
= 0.65 * 0.026 + 0.35 * 0.051
≈ 0.0169 + 0.0179
≈ 0.0348
So, the probability that a randomly selected scooter did not pass inspection is approximately 0.0348, or 3.48%.
b) To find the probability that if a randomly selected scooter passes inspection, it came from assembly line B, we need to use Bayes' theorem.
Let's denote the event "scooter passing inspection" as P and the event "scooter from assembly line B" as B.
We are given that 94.9% of line B's scooters pass inspection, so the probability of a scooter passing inspection if it came from line B, denoted as P(B|P), is 0.949.
We already know the probability of selecting a scooter from line B, denoted as P(B), is 0.35, which we calculated in part a.
Using Bayes' theorem, we can find the probability of a scooter coming from line B if it passes inspection, denoted as P(B|P):
P(B|P) = (P(P|B) * P(B)) / P(P)
= (0.949 * 0.35) / P(P)
To find P(P), we need to use the law of total probability and sum the probabilities of passing inspection for scooters from line A and line B:
P(P) = P(A) * P(P|A) + P(B) * P(P|B)
= 0.65 * 0.974 + 0.35 * 0.949
≈ 0.6331 + 0.3322
≈ 0.9653
Substituting the values back into the equation, we have:
P(B|P) = (0.949 * 0.35) / 0.9653
≈ 0.3309 / 0.9653
≈ 0.3422
So, the probability that if a randomly selected scooter passes inspection, it came from assembly line B is approximately 0.3422, or 34.22%.