A company has three machines A, B and C which all produce the same two parts, X and

Y. of all the parts produced, machine A produces 60%, machine B produces 30%, and
machine C produces the rest. 40% of the parts made by machine A are part X, 50% of the
parts made by machine B are part X, and 70% of the parts made by machine C are part X.
A part produced by this company is randomly sampled and is determined to be an X part.
With the knowledge that it is an X part, find the probabilities that the part came from
machine A, B or C.

A company has three machines A, B and C which all produce the same two parts, X and

Y. of all the parts produced, machine A produces 60%, machine B produces 30%, and
machine C produces the rest. 40% of the parts made by machine A are part X, 50% of the
parts made by machine B are part X, and 70% of the parts made by machine C are part X.
A part produced by this company is randomly sampled and is determined to be an X part.
With the knowledge that it is an X part, find the probabilities that the part came from
machine A, B or C.

Blue Nile University recently conducted a survey of undergraduate students in order to

gather information about the usage of the library. The population for this study included
all 4000 undergraduate students enrolled in the university. The library officers are
interested in increasing usage, particularly among females (F) and seniors (S) at the
university. Of the 4000 students, 800 students are seniors, 1800 students are females and
450 of the 1800 females are seniors.
2 | P a g e L e a d s t a r C o l l e g e o f M a n a g e m e n t a n d L e a d e r s h i p
Required:
 What is the probability that a student selected at random is a senior given that the
selected student is female?
 What is the probability that a student selected at random is female given that the
selected student is senior?

Blue Nile University recently conducted a survey of undergraduate students in order to

gather information about the usage of the library. The population for this study included
all 4000 undergraduate students enrolled in the university. The library officers are
interested in increasing usage, particularly among females (F) and seniors (S) at the
university. Of the 4000 students, 800 students are seniors, 1800 students are females and
450 of the 1800 females are seniors.
2 | P a g e L e a d s t a r C o l l e g e o f M a n a g e m e n t a n d L e a d e r s h i p
Required:
 What is the probability that a student selected at random is a senior given that the
selected student is female?
 What is the probability that a student selected at random is female given that the
selected student is senior?

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To find the probabilities that the X part came from machines A, B, or C, we can use Bayes' theorem. Bayes' theorem allows us to update our probability estimates based on new information.

Let's define the following events:
A: The part came from machine A.
B: The part came from machine B.
C: The part came from machine C.
X: The part is an X part.

We need to find the conditional probabilities P(A|X), P(B|X), and P(C|X) - the probabilities that the part came from each machine given that it is an X part.

According to the problem, we have the following information:
P(A) = 0.6 (probability that a part comes from machine A)
P(B) = 0.3 (probability that a part comes from machine B)
P(C) = 1 - P(A) - P(B) (probability that a part comes from machine C)

P(X|A) = 0.4 (probability that a part from machine A is part X)
P(X|B) = 0.5 (probability that a part from machine B is part X)
P(X|C) = 0.7 (probability that a part from machine C is part X)

We can use Bayes' theorem to find the conditional probabilities:

P(A|X) = (P(X|A) * P(A)) / P(X)
P(B|X) = (P(X|B) * P(B)) / P(X)
P(C|X) = (P(X|C) * P(C)) / P(X)

To find P(X), the probability that a randomly sampled part is an X part, we need to use the law of total probability. Since parts from all three machines can be sampled, we need to calculate the probability of sampling an X part from each machine and then sum them up:

P(X) = P(X|A) * P(A) + P(X|B) * P(B) + P(X|C) * P(C)

Now we can plug in the values and calculate the probabilities.

P(X) = (0.4 * 0.6) + (0.5 * 0.3) + (0.7 * P(C))

Since P(C) = 1 - P(A) - P(B), we can substitute it in the equation:

P(X) = (0.4 * 0.6) + (0.5 * 0.3) + (0.7 * (1 - 0.6 - 0.3))
P(X) = 0.24 + 0.15 + 0.07

To find the probabilities P(A|X), P(B|X), and P(C|X) we substitute the values into the Bayes' theorem formula:

P(A|X) = (0.4 * 0.6) / (0.24 + 0.15 + 0.07)
P(B|X) = (0.5 * 0.3) / (0.24 + 0.15 + 0.07)
P(C|X) = (0.7 * (1 - 0.6 - 0.3)) / (0.24 + 0.15 + 0.07)

Let's calculate these probabilities:

P(A|X) = 0.144 / (0.46)
P(B|X) = 0.15 / (0.46)
P(C|X) = 0.07 / (0.46)

P(A|X) = 0.313
P(B|X) = 0.326
P(C|X) = 0.152

Therefore, the probabilities that the X part came from machines A, B, or C are approximately 0.313, 0.326, and 0.152, respectively.