CFA Latest CFA Level 1 Discussions Can someone help with this question??

Can someone help with this question??

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    • emmanuel.ch
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      I’m having trouble figuring out how to start with some parts of these probability questions, like part A 
      I tried to used P(pass test&non survivor)/ P(non survivor) to get P(passtest|nonsurvivor) 
      But when I checked the solution I saw that they used the unconditional probability of P(pass test) = P(pass test|survivor)* P(survivor) + P(pass test| non survivor)*P(non survivor). 
      I understand why they used that but I don’t understand why the method I chose was wrong, can someone please clarify? How do you know which formula to use in these situations. Thanks, (Question is below) 

      You have developed a set of criteria for evaluating distressed credits. Companies that do not receive a passing

      score are classed as likely to go bankrupt within 12 months. You gathered the following information when
      validating the criteria:
      Forty percent of the companies to which the test is administered will go bankrupt within 12 months: P
      (nonsunivor) = 0.40.
      Fifty-five percent of the companies to which the test is administered pass it: P(pass test)= 0.55.

      • The probability that a company will pass the test given that it will subsequently survive 12 months, is 0.85:
      P(pass test|survivor) =0.85.

      A. What is P (pass test|nonsurvivor)?
      B. Using Bayes’ formula, calculate the probability that a company is a survivor, given that it passes the test;
      that is, calculate P(survivor|pass test).
      C. What is the probability that a company is a nonsurvivor given that it fails the test?
      D. Is this test effective? 
    • PaulAdaptPrep
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      From Jon Lai on the Passed Tense team:

      There is nothing wrong with your approach. You just need to solve the numerator of Pr(pass test & non survivor).

      Consider that you have Pr(non survivor) = 0.4, Pr(pass) = 0.55 and Pr(pass | survivor) = 0.85. You ought to make use of that conditional probability, and it can become Pr(pass & survivor) if you multiply it by Pr(survivor). Therefore, Pr(pass & survivor) = 0.85 * (1 – 0.4) = 0.51.
      So how does Pr(pass), Pr(pass & survivor), and Pr(pass & non survivor) all relate? The law of total probability: Pr(pass) = Pr(pass & survivor) + Pr(pass & non survivor). So you can solve for the numerator as
      Pr(pass & non survivor) = Pr(pass) – Pr(pass & survivor) = 0.55 – 0.51 = 0.04
      Divide 0.04 by 0.4, and there you have it. You’ll notice that what we just did is the same thing as the solution. That formula for Pr(pass test) simply puts all the relationships into one neat expression.
    • emmanuel.ch
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      Many thanks, your explanation was clear. One last question. So I assume P(pass & nonsurvivor) does not = (0.55)*(0.4) because they are not mutually exclusive?? 

      Thanks again

    • PaulAdaptPrep
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      I believe you meant “not independent”. And yes, that’d be correct. You can actually tell (don’t have to assume) that independence does not exist. Pr(pass) = 0.55, while Pr(pass | survivor) = 0.85. If “passing” and “being a survivor” are independent, then – loosely speaking – it wouldn’t make a difference if the event was conditional or not, i.e. Pr( A ) = Pr( A | B ). Since the unconditional probability of passing (0.55) is not the same as the conditional event of passing (0.85), that means both events are not independent.
      Now you might turn around and say: “Well that applies to “passing” and “being a survivor”; how about “passing” and “being a non survivor”?” You have the same deal when it comes to complement of events. So if events A and B are independent, then independence also exists between the pairs A & B’, A’ & B, and A’ & B’.
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