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The probability of A is 0.4. The probability of AC is 0.6. The probability of (B | A) is 0.5, and the probability of (B | AC) is 0.2. Using Bayes’ formula, what is the probability of (A | B)?

A) 0.125.

B) 0.375.

C) 0.625.

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Using the total probability rule, we can compute the P(B):
P(B) = [P(B | A) × P(A)] + [P(B | AC) × P(AC)]
P(B) = [0.5 × 0.4] + [0.2 × 0.6] = 0.32
Using Bayes’ formula, we can solve for P(A | B):
P(A | B) = [ P(B | A) ÷ P(B) ] × P(A) = [0.5 ÷ 0.32] × 0.4 = 0.625

cross-ied

If a firm is going to create three teams of four from twelve employees. Which approach is the most appropriate for determining how the twelve employees can be selected for the three teams?

A) Permutation formula.

B) Combination formula.

C) Labeling formula.

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This problem is a labeling problem where the 12 employees will be assigned one of three labels. It requires the labeling formula.
In this case there are [(12!) / (4!4!4!)] = 34,650 ways to group the employees.

cross-ied

For the task of arranging a given number of items without any sub-groups, this would require:

A) the labeling formula.

B) the permutation formula.

C) only the factorial function.

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The factorial function, denoted n!, tells how many different ways n items can be arranged where all the items are included.

cross-ied

Which of the following statements about counting methods is least accurate?

A) The labeling formula determines the number of different ways to assign a given number of different labels to a set of objects.

B) The combination formula determines the number of different ways a group of objects can be drawn in a specific order from a larger sized group of objects.

C) The multiplication rule of counting is used to determine the number of different ways to choose one object from each of two or more groups.

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The permutation formula is used to find the number of possible ways to draw r objects from a set of n objects when the order in which the objects are drawn matters. The combination formula ("n choose r") is used to find the number of possible ways to draw r objects from a set of n objects when order is not important. The other statements are accurate.

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A supervisor is evaluating ten subordinates for their annual performance reviews. According to a new corporate policy, for every ten employees, two must be evaluated as “exceeds expectations,” seven as “meets expectations,” and one as “does not meet expectations.” How many different ways is it possible for the supervisor to assign these ratings?

A) 10,080.

B) 5,040.

C) 360.

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The number of different ways to assign these labels is:

cross-ied

A firm wants to select a team of five from a group of ten employees. How many ways can the firm compose the team of five?

A) 120.

B) 25.

C) 252.

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This is a labeling problem where there are only two labels: chosen and not chosen. Thus, the combination formula applies: 10! / (5! × 5!) = 3,628,800 / (120 × 120) = 252.With a TI calculator: 10 [2nd][nCr] 5 = 252.

cross-ied

A portfolio manager wants to eliminate four stocks from a portfolio that consists of six stocks. How many ways can the four stocks be sold when the order of the sales is important?

A) 360.

B) 180.

C) 24.

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This is a choose four from six problem where order is important. Thus, it requires the permutation formula: n! / (n − r)! = 6! / (6 − 4)! = 360.
With TI calculator: 6 [2nd][nPr] 4 = 360.

terpsichorefan

thanks for sharing