Reading 8: Probability Concepts-LOS h习题精选 events, center, 2010, interpret

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

LOS h: Calculate and interpret, using the total probability rule, an unconditional probability.

 

 

 

The events Y and Z are mutually exclusive and exhaustive: P(Y) = 0.4 and P(Z) = 0.6. If the probability of X given Y is 0.9, and the probability of X given Z is 0.1, what is the unconditional probability of X?

A) 0.33.

B) 0.42.

C) 0.40.

zaestau

c

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Jay Hamilton, CFA, is analyzing Madison, Inc., a distressed firm. Hamilton believes the firm’s survival over the next year depends on the state of the economy. Hamilton assigns probabilities to four economic growth scenarios and estimates the probability of bankruptcy for Madison under each:

Economic growth scenario	Probability of scenario	Probability of bankruptcy
Recession (< 0%)	20%	60%
Slow growth (0% to 2%)	30%	40%
Normal growth (2% to 4%)	40%	20%
Rapid growth (> 4%)	10%	10%

Based on Hamilton’s estimates, the probability that Madison, Inc. does not go bankrupt in the next year is closest to:

A) 18%.

B) 33%.

C) 67%.

bmaggie

Jay Hamilton, CFA, is analyzing Madison, Inc., a distressed firm. Hamilton believes the firm’s survival over the next year depends on the state of the economy. Hamilton assigns probabilities to four economic growth scenarios and estimates the probability of bankruptcy for Madison under each:

Economic growth scenario	Probability of scenario	Probability of bankruptcy
Recession (< 0%)	20%	60%
Slow growth (0% to 2%)	30%	40%
Normal growth (2% to 4%)	40%	20%
Rapid growth (> 4%)	10%	10%

Based on Hamilton’s estimates, the probability that Madison, Inc. does not go bankrupt in the next year is closest to:

A) 18%.

B) 33%.

C) 67%.

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Using the total probability rule, the unconditional probability of bankruptcy is (0.2)(0.6) + (0.3)(0.4) + (0.4)(0.2) + (0.1) (0.1) = 0.33. The probability that Madison, Inc. does not go bankrupt is 1 – 0.33 = 0.67 = 67%.

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Firm A can fall short, meet, or exceed its earnings forecast. Each of these events is equally likely. Whether firm A increases its dividend will depend upon these outcomes. Respectively, the probabilities of a dividend increase conditional on the firm falling short, meeting or exceeding the forecast are 20%, 30%, and 50%. The unconditional probability of a dividend increase is:

A) 0.333.

B) 0.500.

C) 1.000.

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Firm A can fall short, meet, or exceed its earnings forecast. Each of these events is equally likely. Whether firm A increases its dividend will depend upon these outcomes. Respectively, the probabilities of a dividend increase conditional on the firm falling short, meeting or exceeding the forecast are 20%, 30%, and 50%. The unconditional probability of a dividend increase is:

A) 0.333.

B) 0.500.

C) 1.000.

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The unconditional probability is the weighted average of the conditional probabilities where the weights are the probabilities of the conditions. In this problem the three conditions fall short, meet, or exceed its earnings forecast are all equally likely. Therefore, the unconditional probability is the simple average of the three conditional probabilities: (0.2 + 0.3 + 0.5) ÷ 3.

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The events Y and Z are mutually exclusive and exhaustive: P(Y) = 0.4 and P(Z) = 0.6. If the probability of X given Y is 0.9, and the probability of X given Z is 0.1, what is the unconditional probability of X?

A) 0.33.

B) 0.42.

C) 0.40.

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Because the events are mutually exclusive and exhaustive, the unconditional probability is obtained by taking the sum of the two joint probabilities: P(X) = P(X | Y) × P(Y) + P(X | Z) × P(Z) = 0.4 × 0.9 + 0.6 × 0.1 = 0.42.