(Weighing produce). A supermarket has three employees who package and weigh produce. Employee A records the correct weight 98% of the time. Employees B and C record the correct weight 97% and 95% of the time, respectively. Employees A, B, and C handle 40%, 40% and 20% of the packaging, respectively. A customer complains about the incorrect weight recorded on a package he has purchased. What is the probability that the package was weighed by employee C?
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Math - Damon, Thursday, March 27, 2014 at 11:57am
A wrong .02 * .4 = .008
B wrong .03 * .4 = .012
C wrong .05 * .2 = .01
total wrong = .03
C wrong/total wrong = .01/.03 = .33
To find the probability that the package was weighed by employee C given that there was an incorrect weight recorded, we can use Bayes' Theorem.
Let's define the following events:
A: Employee A weighed the package.
B: Employee B weighed the package.
C: Employee C weighed the package.
I: Incorrect weight recorded.
We are trying to find P(C | I), the probability of employee C weighing the package given that there was an incorrect weight recorded.
We know the following probabilities:
P(A) = 0.40 (Employee A handles 40% of the packaging)
P(B) = 0.40 (Employee B handles 40% of the packaging)
P(C) = 0.20 (Employee C handles 20% of the packaging)
P(I | A) = 0.02 (Employee A records the correct weight 98% of the time)
P(I | B) = 0.03 (Employee B records the correct weight 97% of the time)
P(I | C) = 0.05 (Employee C records the correct weight 95% of the time)
According to Bayes' Theorem:
P(C | I) = (P(I | C) * P(C)) / (P(I | A) * P(A) + P(I | B) * P(B) + P(I | C) * P(C))
Plugging in the values:
P(C | I) = (0.05 * 0.20) / ((0.02 * 0.40) + (0.03 * 0.40) + (0.05 * 0.20))
= 0.01 / (0.008 + 0.012 + 0.01)
= 0.01 / 0.03
= 1/3
= 0.3333
Therefore, the probability that the package was weighed by employee C given that there was an incorrect weight recorded is approximately 0.33 (or 33.33%).
To solve this problem, we need to use Bayes' theorem. Bayes' theorem allows us to calculate the probability of an event occurring given prior knowledge of related events.
Let's define the events:
A: The package was weighed by employee A.
B: The package was weighed by employee B.
C: The package was weighed by employee C.
D: The package had an incorrect weight.
From the given information, we know:
P(A) = 0.4 (Employee A handles 40% of the packaging)
P(B) = 0.4 (Employee B handles 40% of the packaging)
P(C) = 0.2 (Employee C handles 20% of the packaging)
P(D | A) = 0.02 (Employee A records the correct weight 98% of the time, so the probability of an incorrect weight is 1 - 0.98 = 0.02)
P(D | B) = 0.03 (Employee B records the correct weight 97% of the time, so the probability of an incorrect weight is 1 - 0.97 = 0.03)
P(D | C) = 0.05 (Employee C records the correct weight 95% of the time, so the probability of an incorrect weight is 1 - 0.95 = 0.05)
We want to calculate P(C | D), i.e., the probability that the package was weighed by employee C given that the weight recorded was incorrect.
According to Bayes' theorem:
P(C | D) = (P(D | C) * P(C)) / P(D)
To calculate P(D), we need to consider all the possible ways the package could have been weighed incorrectly, regardless of who weighed it:
P(D) = P(D | A) * P(A) + P(D | B) * P(B) + P(D | C) * P(C)
Substituting the given values, we have:
P(D) = (0.02 * 0.4) + (0.03 * 0.4) + (0.05 * 0.2) = 0.008 + 0.012 + 0.01 = 0.03
Now we can calculate P(C | D):
P(C | D) = (0.05 * 0.2) / 0.03 = 0.01 / 0.03 = 0.3333
Therefore, the probability that the package was weighed by employee C given the recorded weight was incorrect is approximately 0.3333 or 33.33%.