Widgets are produced at a certain factory by each of three machines A, B and C. These machines produce 1000, 600 and 400 widgets per day, respectively. The probability that a given widget is defective is 4% for one produced by Machine A, 3% if produced by Machine B, and 2% if produced by Machine C. Suppose that the Widget Inspector selects a widget at random from the factory's widget inventory.
(a) What is the probability that the widget is defective?
(b) If the widget is defective, what is the chance that it came from Machine A?
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Answer a) is it 4+ 3+2 =9%
b) A=widget is defective
B=came from machine A
P(A& B)= P(A)P(B)
= .09* (1000/2000)= .045
= 4.5 %
Hey there! Let me give you a humorous take on this:
(a) So if we add up the probabilities of each machine producing a defective widget, we get 4% + 3% + 2% = 9%. So there's a 9% chance that the widget is defective. Better hope you don't get a lemon!
(b) Now, let's look at the chance that a defective widget came from Machine A. We can use some probability magic here. So, let's call the chance that a widget is defective as "A" and the chance that it came from Machine A as "B". Using some fancy math, we get P(A and B) = P(A) * P(B). In this case, P(A) is 9% and P(B) is 1000/2000 (since Machine A produces 1000 out of the total 2000 widgets). Crunching the numbers, we get 0.045, or 4.5%. So, if you get a defective widget, there's a 4.5% chance it came from Machine A. Those machines are just clowning around, aren't they?
To solve this problem, we need to use the concept of conditional probability. Let's break it down step by step.
(a) To find the probability that the widget is defective, we need to consider the probabilities of the widget being produced by each machine and being defective. We can find the probability of a widget being defective by each machine by multiplying the probability of the widget being produced by that machine with the probability of a widget from that machine being defective.
Probability of a widget being produced by Machine A = 1000 widgets / (1000 + 600 + 400) = 0.5
Probability of a widget from Machine A being defective = 4% = 0.04
Probability of a widget being produced by Machine B = 600 / (1000 + 600 + 400) = 0.3
Probability of a widget from Machine B being defective = 3% = 0.03
Probability of a widget being produced by Machine C = 400 / (1000 + 600 + 400) = 0.2
Probability of a widget from Machine C being defective = 2% = 0.02
Now, we can calculate the overall probability of a widget being defective by summing up the individual probabilities:
Probability of a widget being defective = (0.5 * 0.04) + (0.3 * 0.03) + (0.2 * 0.02)
= 0.02 + 0.009 + 0.004
= 0.033 or 3.3%
So, the probability that the widget is defective is 3.3%.
(b) To find the chance that the defective widget came from Machine A, we need to use conditional probability. We know that the widget is defective, and we want to find the probability that it came from Machine A. We can use the formula:
P(A|B) = P(A ∩ B) / P(B)
Let's define the events:
A: The widget is defective
B: The widget came from Machine A
We know the probabilities:
P(A) = 0.033 (from part a)
P(B) = 0.5 (probability of a widget being produced by Machine A)
And we can calculate P(A ∩ B) by multiplying the probability of a widget being produced by Machine A with the probability of a widget from Machine A being defective:
P(A ∩ B) = P(A) * P(B) = 0.033 * 0.5 = 0.0165 or 1.65%
Now, we can calculate the probability that the defective widget came from Machine A:
P(A|B) = P(A ∩ B) / P(B) = (0.0165 / 0.5) * 100% = 0.033 * 100% = 3.3%
So, the chance that the defective widget came from Machine A is 3.3%.