Electric motors coming off two assembly lines are pooled for storage in a common stockroom, and the room contains an equal number of motors from each line. Motors are periodically sampled from that room and tested. It is known that 10% of the motors from line I are defective and 15% of the motors from line II are defective. If a motor is randomly selected from the stockroom and found to be defective, find the probability that it came from line I.
0.4
Ah, the world of defective motors! A truly electrifying topic, if I do say so myself. So, we have two assembly lines, I and II, both contributing motors to a common stockroom. Fascinating!
Let's break it down. We know that 10% of the motors from line I are defective and 15% of the motors from line II are defective. Quite shocking, isn't it?
Now, if a motor is randomly selected from the stockroom and is found to be defective, we want to find the probability that it came from line I. This calls for a bit of mathematical clownery!
Let's assume that there are an equal number of motors from each line in the stockroom. Since line I motors make up 50% of the sample (50% from line I and 50% from line II), it seems logical to assume that the probability of picking a defective motor from line I is higher than from line II. But how much higher?
To calculate the probability, we need to consider both the overall probability of picking a defective motor and the individual probabilities for each line.
The overall probability of picking a defective motor can be calculated by summing up the probabilities from each line:
(0.10 * 0.50) + (0.15 * 0.50) = 0.05 + 0.075 = 0.125
Now, let's determine the probability specifically for line I. We want to find the proportion of defective motors that come from line I:
(0.10 * 0.50) / 0.125 = 0.05 / 0.125 = 0.4
So, the probability that a randomly selected defective motor comes from line I is 0.4, or 40%.
In essence, there's a 40% chance that the motor causing all the trouble is from line I. Remember, probabilistic clowning is an amusing game, but also quite useful in finding answers!
To find the probability that a randomly selected defective motor came from line I, we can use Bayes' theorem.
Let's define the events:
A: The motor comes from line I
B: The motor is defective
We need to find P(A|B), which represents the probability that the motor comes from line I given that it is defective.
According to Bayes' theorem:
P(A|B) = (P(B|A) * P(A))/P(B)
P(B|A) = 0.10 (the probability that a motor is defective given that it comes from line I)
P(A) = 0.50 (since there is an equal number of motors from each line in the stockroom)
P(B) = P(B|A) * P(A) + P(B|¬A) * P(¬A)
P(B|¬A) = 0.15 (the probability that a motor is defective given that it doesn't come from line I)
P(¬A) = 0.50 (since there is an equal number of motors from each line in the stockroom)
Substituting these values into the equation:
P(A|B) = (0.10 * 0.50) / (0.10 * 0.50 + 0.15 * 0.50)
Simplifying further:
P(A|B) = 0.05 / (0.05 + 0.075)
Calculating:
P(A|B) = 0.05 / 0.125
P(A|B) ≈ 0.4
Therefore, the probability that the randomly selected defective motor came from line I is approximately 0.4 or 40%.
To solve this problem, we can use Bayes' Theorem, which relates conditional probabilities.
Let's define the following events:
- Event A: The motor is defective.
- Event B: The motor came from line I.
We need to find the probability of Event B given Event A, denoted as P(B|A). We can calculate this using Bayes' Theorem:
P(B|A) = P(A|B) * P(B) / P(A)
We can break down the calculation step by step:
1. P(A|B): This is the probability of the motor being defective, given that it came from line I. We are given that 10% of the motors from line I are defective, so P(A|B) = 0.10.
2. P(B): This is the probability that a randomly selected motor came from line I. Since both assembly lines have an equal number of motors in the stockroom, P(B) = 0.5 (50%).
3. P(A): This is the probability of a randomly selected motor being defective, regardless of which line it came from. We can calculate this using the law of total probability:
P(A) = P(A|B) * P(B) + P(A|not B) * P(not B)
Given that line II has 15% defective motors, and both assembly lines contribute equally to the stockroom, we have:
P(A) = P(A|B) * P(B) + P(A|not B) * P(not B)
= 0.10 * 0.5 + 0.15 * 0.5
= 0.05 + 0.075
= 0.125 (12.5%)
Now we can calculate P(B|A) using Bayes' Theorem:
P(B|A) = P(A|B) * P(B) / P(A)
= 0.10 * 0.5 / 0.125
= 0.05 / 0.125
= 0.4 (40%)
Therefore, the probability that a randomly selected defective motor came from line I is 40%.