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?
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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.
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:
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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.
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:
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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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