Reading 8: Probability Concepts-LOS f 习题精选 2011, interpret, number, center

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

LOS f: Calculate and interpret 1) the joint probability of two events, 2) the probability that at least one of two events will occur, given the probability of each and the joint probability of the two events, and 3) a joint probability of any number of independent 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) 28%.

B) 48%.

C) 25%.

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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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If two fair coins are flipped and two fair six-sided dice are rolled, all at the same time, what is the probability of ending up with two heads (on the coins) and two sixes (on the dice)?

A) 0.0069.

B) 0.8333.

C) 0.4167.

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For the four independent events defined here, the probability of the specified outcome is 0.5000 × 0.5000 × 0.1667 × 0.1667 = 0.0069.

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If two events are independent, the probability that they both will occur is:

A) 0.50.

B) Cannot be determined from the information given.

C) 0.00.

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If two events are independent, their probability of their joint occurrence is computed as follows:  P(A∩B) = P(A) × P(B). Since we are not given any information on the respective probabilities of A or B, there is not enough information.

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A very large company has twice as many male employees relative to female employees. If a random sample of four employees is selected, what is the probability that all four employees selected are female?

A) 0.0123.

B) 0.3333.

C) 0.0625.

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Since there are twice as many male employees to female employees, P(male) = 2/3 and P(female) = 1/3. Therefore, the probability of 4 “successes” = (0.333)⁴ = 0.0123.

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There is a 30% probability of rain this afternoon. There is a 10% probability of having an umbrella if it rains. What is the chance of it raining and having an umbrella?

A) 3%.

B) 40%.

C) 33%.

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P(A) = 0.30. P(B | A) = 0.10. P(AB) = (0.30)(0.10) = 0.03 or 3%.

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A bond portfolio consists of four BB-rated bonds. Each has a probability of default of 24% and these probabilities are independent. What are the probabilities of all the bonds defaulting and the probability of all the bonds not defaulting, respectively?

A) 0.96000; 0.04000.

B) 0.04000; 0.96000.

C) 0.00332; 0.33360.

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For the four independent events where the probability is the same for each, the probability of all defaulting is (0.24)⁴. The probability of all not defaulting is (1 ? 0.24)⁴.

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A very large company has equal amounts of male and female employees. If a random sample of four employees is selected, what is the probability that all four employees selected are female?

A) 0.0256

B) 0.1600

C) 0.0625.

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Each employee has equal chance of being male or female. Hence, probability of 4 “successes” = (0.5)⁴ = 0.0625

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Which of the following is a joint probability? The probability that a:

A) stock pays a dividend and splits next year.

B) company merges with another firm next year.

C) stock increases in value after an increase in interest rates has occurred.

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A joint probability applies to two events that both must occur.

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Jessica Fassler, options trader, recently wrote two put options on two different underlying stocks (AlphaDog Software and OmegaWolf Publishing), both with a strike price of $11.50. The probabilities that the prices of AlphaDog and OmegaWolf stock will decline below the strike price are 65% and 47%, respectively. The probability that at least one of the put options will fall below the strike price is approximately:

A) 0.31.

B) 1.00.

C) 0.81.

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We calculate the probability that at least one of the options will fall below the strike price using the addition rule for probabilities (A represents AlphaDog, O represents OmegaWolf):

P(A or O) = P(A) + P(O) ? P(A and O), where P(A and O) = P(A) × P(O)
P(A or O) = 0.65 + 0.47 ? (0.65 × 0.47) = approximately 0.81

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Thomas Baynes has applied to both Harvard and Yale. Baynes has determined that the probability of getting into Harvard is 25% and the probability of getting into Yale (his father’s alma mater) is 42%. Baynes has also determined that the probability of being accepted at both schools is 2.8%. What is the probability of Baynes being accepted at either Harvard or Yale, but not both?

A) 7.7%.

B) 10.5%.

C) 64.2%.

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Using the addition rule, the probability of being accepted at Harvard or Yale, but not both, is equal to: P(Harvard) + P(Yale) ? P(Harvard and Yale) = 0.25 + 0.42 ? 0.028 = 0.642 or 64.2%.