A foreman for an injection-molding firm admits that on 46% of his shifts, he forgets to shut off the injection machine on his line. This causes the machine to overheat, increasing the probability that a defective molding will be produced during the early morning run from 5% to 19%. If a molding is randomly selected from the early morning run of a random day, what is the probability that it is defective?
I got:
P(F and D) = 0.46*0.19 = 0.0874
P(~F and D) = 0.54*0.05 = 0.027
P(F|D) = P(F and D) / [P(F and D) + P(~F and D)] = 0.0874 / 0.1144 = 0.76398601398
But my online homework's saying the answers incorrect
P(FD) = .46
P(notFD) = . 54
If he forgets, then increase is P(D) = .19
If doesn’t forget, remain P(D) .05
Therefore, P(FD)*P(D) +P(NotFD)*P(D)
(.46*.19) +(.54*.05) =.1144
Well, well, well... looks like someone's having a "defective" moment with their answer! Let's inject some humor into this problem and figure it out.
First, let's break it down for you. You correctly calculated the probability of the foreman forgetting to shut off the injection machine and a defective molding being produced (P(F and D) = 0.46 * 0.19 = 0.0874). Kudos to you!
But now, we need to consider the probability of the injection machine being properly shut off and a defective molding still being produced (P(~F and D) = 0.54 * 0.05 = 0.027).
To find the overall probability of getting a defective molding, we need to calculate the probability of the molding being defective, given that it was produced during the early morning run. So let's do some clown calculations:
P(D) = P(F and D) + P(~F and D)
= 0.0874 + 0.027
= 0.1144
Now, to calculate the probability that a randomly selected molding is defective, we can use Bayes' Theorem:
P(D|F) = P(F and D) / P(F)
= 0.0874 / 0.46
≈ 0.19
So, the probability that a randomly selected molding from the early morning run is defective is approximately 0.19 or 19%.
I hope this clears up any confusion, and remember, humor is always there to keep us entertained, even in probability problems!
Your calculations are correct up to this point:
P(F and D) = 0.46 * 0.19 = 0.0874
P(~F and D) = 0.54 * 0.05 = 0.027
However, to find the probability that a randomly selected molding is defective, you need to consider both the cases where the foreman forgets to shut off the machine (F) and when he doesn't forget (not F). The probability of a defective molding given the foreman forgets is 0.19, and the probability of a defective molding given the foreman doesn't forget is 0.05.
To calculate the overall probability, you need to consider the probabilities of both cases happening:
P(D) = P(F and D) + P(~F and D)
Plugging in the values:
P(D) = 0.0874 + 0.027 = 0.1144
So the probability that a randomly selected molding is defective is 0.1144, or 11.44%.
To calculate the probability that a molding is defective, we need to consider two scenarios: when the foreman forgets to shut off the injection machine (F) and when he doesn't forget (¬F).
Let's break down the steps to find the correct answer:
Step 1: Calculate the overall probability of the foreman forgetting (F) and not forgetting (¬F). According to the information given, the foreman forgets on 46% of his shifts, so:
P(F) = 0.46
P(¬F) = 1 - P(F) = 0.54
Step 2: Calculate the probability of producing a defective molding given that the machine was not shut off (D|¬F) and given that the machine was shut off (D|F). From the information provided, when the machine is not shut off, the probability of a defective molding is 19%, and when it is shut off, the probability is 5%:
P(D|¬F) = 0.19
P(D|F) = 0.05
Step 3: Calculate the probability of selecting a defective molding (D) by applying the Law of Total Probability:
P(D) = P(D|¬F) * P(¬F) + P(D|F) * P(F)
Substituting in the values we calculated:
P(D) = 0.19 * 0.54 + 0.05 * 0.46
= 0.1026 + 0.023
= 0.1256
Therefore, the correct probability that a randomly selected molding from the early morning run is defective is approximately 0.1256 or 12.56%.