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The following two questions are based on the wondrous Statistical topic probability. After attempting both questions I have yet to answer either correctly. If anyone has encountered similar problems before and understands how each is answered, I would be wholly appreciative if you could provide a systematic breakdown of how to do either, or both if possible.

  1. A publisher sends advertising materials for an accounting text to 80% of all professors teaching the appropriate accounting course. Thirty percent of the professors who received this material adopted the book, as did 10% of the professors who did not receive the material. What is the probability that a professor who adopts the book has received the advertising material?

  2. A stock market analyst examined the prospects of the shares of a large number of corporations when the performance of these stocks was investigated one year later, it turned out that 25% performed much better that the market average, 25%, much worse and the remaining 50%, about the same as the average. Forty percent of the stocks that turned out to do much better than the market were rated good buys by the analyst, as were 20% of those that did about as well as the market and 10% of those that did much worse. What is the probability that a stock rated a good buy by the analyst performed much better than the average?

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Both are exercises in conditional probability. Let's do the first one together and you can do the second one in the same way then.

You have 2 classifications - sent/not-sent (S/NS for short) and adopted/not-adopted (A/NA). That makes your underlying population split into 4 categories: (S,A), (S,NA), (NS,A), (NS,NA). First, using the data in the book, figure out the percentages corresponding to each. (E.g. that he sent it to 80% tells you that $(S,A) + (S,NA) = 0.8$ and $(NS,A)+(NA,NA) = 1-.8 = .2$. Figure out exactly what is the value of each of the 4 sub-classes.)

Then, you are looking for $$ \mathbb{P}[S|A] = \frac{\mathbb{P}[S \text{ and } A]}{\mathbb{P}[A]} = \frac{(S,A)}{(S,A) + (NS,A)}. $$

Can you take it from here?

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  • $\begingroup$ That is a brilliant explanation, thank you! So for future reference, when presented with a question on this particular facet of conditional probability, I take the following steps: 1. identify the "classifications" in the question; 2. ascertain the corresponding probability of each; 3. and then use that formulae above? I found the answer after substituting the probabilities into the formulae to be 4/5 or alternatively 0.8. Can you confirm whether this is correct or not? Further, do you know much about Bayes' theorem as I believe that to be pertinent to the second question? $\endgroup$ – Scott Goddard Apr 9 '14 at 19:11
  • $\begingroup$ @ScottGoddard Looks like you evaluated incorrectly. What numbers did you get for each of the 4 sets, and how did you get them? $\endgroup$ – gt6989b Apr 9 '14 at 19:17
  • $\begingroup$ From the explanation you provided I figured (S,A) equates to 0.8, but I have just realised my mistake: I didn't divide the 0.8 by two to divulge what probability (S,A) is on itself. I now acknowledge the probability of (S,A) to be 0.4 (0.8/2). When I input this value into the formula, I have 0.4 as the numerator and then 0.4 + 0.1 as the denominator. I believe this to be the case because (NS, A) + (NA, NA) equals 0.2, so to calculate (NS, A) is 0.2 / 2, so 0.1. If you can highlight at which point or points I've made mistakes, I can then gain an understanding of how to do it correctly. $\endgroup$ – Scott Goddard Apr 9 '14 at 19:48
  • $\begingroup$ @ScottGoddard but the problem says that "$30\%$ of the professors who received this material adopted the book", so $(S,A) = 0.3 \times 0.8 = 0.24$! $\endgroup$ – gt6989b Apr 9 '14 at 20:00

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