mastery 3

1.

An analyst thinks that next year there is a 10% chance that the world economy will be good, a 10% chance that it will be neutral, and a 80% chance that it will be poor. She also predicts probabilities that the performance of a start-up firm, Creative Ideas, will be good, neutral, or poor for each of the economic states of the world economy. The following table presents probabilities for three states of the world economy and the corresponding conditional probabilities for Creative Ideas.

State of

the World

Economy Probability

of Economic

State Performance

of Creative

Ideas Conditional

Probability of

Creative Ideas

Good 0.10 Good 0.60

Neutral 0.10

Poor 0.30

Neutral 0.10 Good 0.10

Neutral 0.10

Poor 0.80

Poor 0.80 Good 0.40

Neutral 0.30

Poor 0.30

PictureClick here for the Excel Data File

a.

What is the probability that the performance of the world economy will be neutral and that of creative ideas will be poor? (Round your answer to 2 decimal places.)

Probability

b.

What is the probability that the performance of Creative Ideas will be poor? (Round your answer to 2 decimal places.)

Probability

c.

The performance of Creative Ideas was poor. What is the probability that the performance of the world economy had also been poor? (Round your answer to 2 decimal places.)

Probability

rev: 08_06_2013_QC_32707

2.

Determine whether the following probabilities are best categorized as subjective, empirical, or classical probabilities.

a.

Before flipping a fair coin, Sunil assesses that he has a 50% chance of obtaining tails.

Subjective probability

Empirical probability

Classical probability

b.

At the beginning of the semester, John believes he has a 90% chance of receiving straight A’s.

Subjective probability

Empirical probability

Classical probability

c.

A political reporter announces that there is a 48% chance that the next person to come out of the conference room will be a Republican, since there are 85 Republicans and 91 Democrats in the room.

Subjective probability

Empirical probability

Classical probability

3.

Christine has always been weak in mathematics. Based on her performance prior to the final exam in Calculus, there is a 39% chance that she will fail the course if she does not have a tutor. With a tutor, her probability of failing decreases to 9%. There is only a 49% chance that she will find a tutor at such short notice.

a.

What is the probability that Christine fails the course? (Round your answer to 4 decimal places.)

Probability

b.

Christine ends up failing the course. What is the probability that she had found a tutor? (Round your answer to 4 decimal places.)

Probability

rev: 08_06_2013_QC_32707

4.

A professor of management has heard that nine students in his class of 49 have landed an internship for the summer. Suppose he runs into two of his students in the corridor.

a.

Find the probability that neither of these students has landed an internship. (Round your intermediate calculations and final answer to 4 decimal places.)

formula176.mml

b.

Find the probability that both of these students have landed an internship. (Round your intermediate calculations and final answer to 4 decimal places.)

P(T1 ∩ T2)

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5.

Consider the following probabilities: P(Ac) = 0.32, P(B) = 0.58, and P(A ∩ Bc) = 0.25.

a. Find P(A | Bc). (Do not round intermediate calculations. Round your answer to 2 decimal places.)

P(A | Bc)

b. Find P(Bc | A). (Do not round intermediate calculations. Round your answer to 3 decimal places.)

P(Bc | A)

c. Are A and B independent events?

Yes because P(A | Bc) = P(A).

Yes because P(A ∩ Bc) ≠ 0.

No because P(A | Bc) ≠ P(A).

No because P(A ∩ Bc) ≠ 0.

rev: 08_06_2013_QC_32707

6.

The State Police are trying to crack down on speeding on a particular portion of the Massachusetts Turnpike. To aid in this pursuit, they have purchased a new radar gun that promises greater consistency and reliability. Specifically, the gun advertises ± one-mile-per-hour accuracy 85% of the time; that is, there is a 0.85 probability that the gun will detect a speeder, if the driver is actually speeding. Assume there is a 2% chance that the gun erroneously detects a speeder even when the driver is below the speed limit. Suppose that 82% of the drivers drive below the speed limit on this stretch of the Massachusetts Turnpike.

a.

What is the probability that the gun detects speeding and the driver was speeding? (Round your answer to 4 decimal places.)

Probability

b.

What is the probability that the gun detects speeding and the driver was not speeding? (Round your answer to 4 decimal places.)

Probability

c.

Suppose the police stop a driver because the gun detects speeding. What is the probability that the driver was actually driving below the speed limit? (Round your answer to 4 decimal places.)

Probability

rev: 08_06_2013_QC_32707

7.

The probabilities that stock A will rise in price is 0.50 and that stock B will rise in price is 0.50. Further, if stock B rises in price, the probability that stock A will also rise in price is 0.35.

a.

What is the probability that at least one of the stocks will rise in price? (Round your answer to 2 decimal places.)

Probability

b. Are events A and B mutually exclusive?

Yes because P(A | B) = P(A).

Yes because P(A ∩ B) = 0.

No because P(A | B) ≠ P(A).

No because P(A ∩ B) ≠ 0.

c. Are events A and B independent?

Yes because P(A | B) = P(A).

Yes because P(A ∩ B) = 0.

No because P(A | B) ≠ P(A).

No because P(A ∩ B) ≠ 0.

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8.

Consider the following joint probability table.

B1 B2 B3 B4

A 0.10 0.09 0.10 0.12

Ac 0.16 0.20 0.09 0.14

PictureClick here for the Excel Data File

a. What is the probability that A occurs? (Round your answer to 2 decimal places.)

Probability

b. What is the probability that B2 occurs? (Round your answer to 2 decimal places.)

Probability

c. What is the probability that Ac and B4 occur? (Round your answer to 2 decimal places.)

Probability

d. What is the probability that A or B3 occurs? (Round your answer to 2 decimal places.)

Probability

e.

Given that B2 has occurred, what is the probability that A occurs? (Round your intermediate calculations and final answers to 4 decimal places.)

Probability

f.

Given that A has occurred, what is the probability that B4 occurs? (Round your intermediate calculations and final answers to 4 decimal places.)

Probability

rev: 08_06_2013_QC_32707

9.

Let P(A) = 0.66, P(B) = 0.31, and P(A ∩ B) = 0.21.

a. Calculate P(A | B). (Round your answer to 2 decimal places.)

P(A | B)

b. Calculate P(A U B). (Round your answer to 2 decimal places.)

P(A U B)

c. Calculate P((A U B)c). (Round your answer to 2 decimal places.)

P((A U B)c)

rev: 08_06_2013_QC_32707

10.

At a local bar in a small Midwestern town, beer and wine are the only two alcoholic options. The manager noted that of all male customers who visited over the weekend, 160 ordered beer, 33 ordered wine, and 35 asked for soft drinks. Of female customers, 46 ordered beer, 24 ordered wine, and 17 asked for soft drinks.

a.

Construct a contingency table that shows frequencies for the qualitative variables Gender (male or female) and Drink Choice (beer, wine, or soft drink).

Drink Choice

Gender Beer (B) Wine (W) Soft Drinks (D) Totals

Male (M)

Female (F)

Total

b. Find the probability that a customer orders wine. (Round your intermediate calculations and final answer to 4 decimal places.)

P(W)

c.

What is the probability that a male customer orders wine? (Round your intermediate calculations and final answer to 4 decimal places.)

P (W | M )

d. Are the events “Wine” and “Male” independent?

Yes because P(“Wine” | “Male”) = P(“Wine”).

Yes because P(“Wine” ∩ “Male”) = P(“Wine”).

No because P(“Wine” | “Male”) ≠ P(“Wine”).

No because P(“Wine” ∩ “Male”) ≠ P(“Wine”).

11.

Let P(A) = 0.42, P(B) = 0.37, and P(A ∩ B) = 0.24.

a. Are A and B independent events?

Yes because P(A | B) = P(A).

Yes because P(A ∩ B) ≠ 0.

No because P(A | B) ≠ P(A).

No because P(A ∩ B) ≠ 0.

b. Are A and B mutually exclusive events?

Yes because P(A | B) = P(A).

Yes because P(A ∩ B) ≠ 0.

No because P(A | B) ≠ P(A).

No because P(A ∩ B) ≠ 0.

c. What is the probability that neither A nor B takes place? (Round your answer to 2 decimal places.)

Probability

rev: 08_06_2013_QC_32707

12.

Records show that 12% of all college students are foreign students who also smoke. It is also known that 60% of all foreign college students smoke. What percent of the students at this university are foreign?

Percent of the students %

13.

Let P(A) = 0.49, P(B | A) = 0.34, and P(B | Ac) = 0.10. Use a probability tree to calculate the following probabilities: (Round your answers to 3 decimal places.)

a. P(Ac)

b. P(A ∩ B)

P(Ac ∩ B)

c. P(B)

d. P(A | B)

14.

Complete the following probability table. (Round Prior Probability answers to 2 decimal places and intermediate calculations and other answers to 4 decimal places.)

Prior

Probability Conditional Probability Joint

Probability Posterior

Probability

P(B) 0.53 P(A | B) 0.15 P(A ∩ B ) P(B | A)

P(Bc) P(A | Bc) 0.38 P(A ∩ Bc) P(Bc | A)

Total P(A) Total

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15.

Consider the following contingency table.

B Bc

A 22 26

Ac 25 27

a.

Convert the contingency table into a joint probability table. (Round your intermediate calculations and final answers to 4 decimal places.)

B

Bc

Total

A

Ac

Total

b. What is the probability that A occurs? (Round your intermediate calculations and final answer to 4 decimal places.)

Probability

c. What is the probability that A and B occur? (Round your intermediate calculations and final answer to 4 decimal places.)

Probability

d.

Given that B has occurred, what is the probability that A occurs? (Round your intermediate calculations and final answer to 4 decimal places.)

Probability

e.

Given that Ac has occurred, what is the probability that B occurs? (Round your intermediate calculations and final answer to 4 decimal places.)

Probability

f. Are A and B mutually exclusive events?

Yes because P(A | B) ≠ P(A).

Yes because P(A ∩ B) ≠ 0.

No because P(A | B) ≠ P(A).

No because P(A ∩ B) ≠ 0.

g. Are A and B independent events?

Yes because P(A | B) ≠ P(A).

Yes because P(A ∩ B) ≠ 0.

No because P(A | B) ≠ P(A).

No because P(A ∩ B) ≠ 0.

rev: 08_06_2013_QC_32707, 11_10_2013_QC_38348

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