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Nate (similar to Freddy, who was discussed in class) is trying to decide whether Kimaya is a nice person. There is an equal chance that Kimaya is mean, average, or nice. Whenever a person meets Kimaya, she acts either pleasantly or unpleasantly. If Kimaya is mean, she will act pleasantly one-quarter of the time; if she is average, she will act pleasantly one-half of the time; and if she is nice, she will act pleasantly three-quarters of the time (everyone has their bad days). In reality, Kimaya’s behavior is independent over time. But Nate believes in the law of small numbers, and has N = 24.
(a) Suppose Nate meets Kimaya four times, and she is pleasant each time. This part asks you to calculate the correct conclusions Nate should draw from this.
Calculate the true probability that Kimaya is pleasant four times in a row if she is mean.
Hint: If they meet only once, what is the true probability that Kimaya is pleasant that time if she is mean? If they meet twice, what is the true probability that she is mean both times if she is mean? If they meet three times, what is the true probability that she is mean all three times?
Calculate the true probability that Kimaya is pleasant four times in a row if she is average.
Calculate the true probability that Kimaya is pleasant four times in a row if she is nice.
Based on the answers to the previous parts, calculate the overall probability that Kimaya is pleasant four times in a row.
Apply Bayes’ rule and calculate the probability that Kimaya is nice conditional on the fact that she was pleasant four times in a row. That is to say, if Nate has seen Kimaya be pleasant all four times, what is the probability that she is truly nice?
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