Reading 8: Probability Concepts-LOS f, (Part 1)习题精选 2010, interpret, parking, center

bmaggie

Session 2: Quantitative Methods: Basic Concepts
Reading 8: Probability Concepts

LOS f, (Part 1): Calculate and interpret the joint probability of two events.

 

 

 

A parking lot has 100 red and blue cars in it.

• 40% of the cars are red.
• 70% of the red cars have radios.
• 80% of the blue cars have radios.

 

What is the probability of selecting a car at random and having it be red and have a radio?

A) 48%.

B) 25%.

C) 28%.

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Joint probability is the probability that both events, in this case a car being red and having a radio, happen at the same time. Joint probability is computed by multiplying the individual event probabilities together: P(red and radio) = (P(red)) × (P(radio)) = (0.4) × (0.7) = 0.28 or 28%.

	Radio	No Radio
Red	28	12	40
Blue	48	12	60
	76	24	100

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What is the probability of selecting a car at random that is either red or has a radio?

A) 76%.

B) 28%.

C) 88%.

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The addition rule for probabilities is used to determine the probability of at least one event among two or more events occurring, in this case a car being red or having a radio. To use the addition rule, the probabilities of each individual event are added together, and, if the events are not mutually exclusive, the joint probability of both events occurring at the same time is subtracted out: P(red or radio) = P(red) + P(radio) ? P(red and radio) = 0.40 + 0.76 ? 0.28 = 0.88 or 88%.

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What is the probability that the car is red given that you already know that it has a radio?

A) 47%.

B) 37%.

C) 28%.

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Given a set of prior probabilities for an event of interest, Bayes’ formula is used to update the probability of the event, in this case that the car we already know has a radio is red. Bayes’ formula says to divide the Probability of New Information given Event by the Unconditional Probability of New Information and multiply that result by the Prior Probability of the Event. In this case, P(red car has a radio) = 0.70 is divided by 0.76 (which is the Unconditional Probability of a car having a radio (40% are red of which 70% have radios) plus (60% are blue of which 80% have radios) or ((0.40) × (0.70)) + ((0.60) × (0.80)) = 0.76.) This result is then multiplied by the Prior Probability of a car being red, 0.40. The result is (0.70 / 0.76) × (0.40) = 0.37 or 37%.

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bmaggie

In a given portfolio, half of the stocks have a beta greater than one. Of those with a beta greater than one, a third are in a computer-related business. What is the probability of a randomly drawn stock from the portfolio having both a beta greater than one and being in a computer-related business?

A) 0.667.

B) 0.333.

C) 0.167.

bmaggie

In a given portfolio, half of the stocks have a beta greater than one. Of those with a beta greater than one, a third are in a computer-related business. What is the probability of a randomly drawn stock from the portfolio having both a beta greater than one and being in a computer-related business?

A) 0.667.

B) 0.333.

C) 0.167.

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This is a joint probability. From the information: P(beta > 1) = 0.500 and P(comp. stock | beta > 1) = 0.333. Thus, the joint probability is the product of these two probabilities: (0.500) × (0.333) = 0.167.

bmaggie

Data shows that 75 out of 100 tourists who visit New York City visit the Empire State Building. It rains or snows in New York City one day in five. What is the joint probability that a randomly choosen tourist visits the Empire State Building on a day when it neither rains nor snows?

A) 60%.

B) 95%.

C) 15%.

bmaggie

Data shows that 75 out of 100 tourists who visit New York City visit the Empire State Building. It rains or snows in New York City one day in five. What is the joint probability that a randomly choosen tourist visits the Empire State Building on a day when it neither rains nor snows?

A) 60%.

B) 95%.

C) 15%.

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A joint probability is the probability that two events occur when neither is certain or a given. Joint probability is calculated by multiplying the probability of each event together. (0.75) × (0.80) = 0.60 or 60%.

bmaggie

Helen Pedersen has all her money invested in either of two mutual funds (A and B). She knows that there is a 40% probability that fund A will rise in price and a 60% chance that fund B will rise in price if fund A rises in price. What is the probability that both fund A and fund B will rise in price?

A) 0.24.

B) 1.00.

C) 0.40.

bmaggie

Helen Pedersen has all her money invested in either of two mutual funds (A and B). She knows that there is a 40% probability that fund A will rise in price and a 60% chance that fund B will rise in price if fund A rises in price. What is the probability that both fund A and fund B will rise in price?

A) 0.24.

B) 1.00.

C) 0.40.

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P(A) = 0.40, P(B|A) = 0.60. Therefore, P(A?B) = P(A)P(B|A) = 0.40(0.60) = 0.24.

bmaggie

If the probability of both a new Wal-Mart and a new Wendy’s being built next month is 68% and the probability of a new Wal-Mart being built is 85%, what is the probability of a new Wendy’s being built if a new Wal-Mart is built?

A) 0.60.

B) 0.70.

C) 0.80.

bmaggie

If the probability of both a new Wal-Mart and a new Wendy’s being built next month is 68% and the probability of a new Wal-Mart being built is 85%, what is the probability of a new Wendy’s being built if a new Wal-Mart is built?

A) 0.60.

B) 0.70.

C) 0.80.

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P(AB) = P(A|B) × P(B)

0.68 / 0.85 = 0.80

bmaggie

The probability of a new Wal-Mart being built in town is 64%. If Wal-Mart comes to town, the probability of a new Wendy’s restaurant being built is 90%. What is the probability of a new Wal-Mart and a new Wendy’s restaurant being built?

A) 0.675.

B) 0.576.

C) 0.306.