Am really struggling to get my head around the total probability rule questions, could someone kindly explain a simple method of how to deal with questions like the one below.
Thank you so much.
An analyst develops a set of criteria for evaluating distressed credits. Companies that do not receive a passing score are classed as likely to go bankrupt within the next year. The analyst concludes the following:
· Forty percent of the companies tested will go bankrupt within a year: P(non-survivor) = 0.40.
· Fifty-five percent of companies tested will pass: P(pass test) = 0.55.
· There is an eighty-five percent probability that a company will pass the test given that it survives a year: P(pass test │survivor) = 0.85.
Using the total probability rule, the probability that a company passed the test given that it goes bankrupt can be determined. The P(pass test│nonsurvivor) is closest to?
We know that a company can pass under two scenarios:
Pass given that it survives a year
Pass given that it does not survive a year
Given that these two cases exhaust the entire realm of possible scenarios where the company can pass, these two have to add up to the total probability that this event will happen. Thus:
P(pass test │survivor)P(survivor)+ P(pass test │non-survivor) P(non-survivor)=P(pass test)
0.85* 0.6+ P(pass test │non-survivor) *0.4=0.55
P(pass test │non-survivor)=0.1
Here is a way, that might help memorize the total probability rule:
Let’s say we want to know the probability that you pass the exam. Now assume you have only two choices, either you prepare for the exam or you don’t. If you prepare for the exam, your chance of passing is 90%. Say if you don’t prepare, your chance is 50%. Both options are exhaustive (there is not half hearted preparing, you either prepare or you don’t). Now in order to figure out your total chance of passing, we need to know how likely it is that you prepare. Let’s say you have no other obligations (no work, no family, no life etc.), so your probabiliy of preparing is very high, nameley 99%. Thus we have:
P(pass CFA │prepare)P(prepare)+ P(passCFA │no-prepare) P(no-prepare)=P(pass CFA)
0.9 * 0.99+ 0.5*0.01=0.896
Notice also, as you increase the probability of not preparing, your chances decrease obviously. You can think of it as a weighting.
Imagine you didn’t know about the total probability rule and this question was given to you in high school. How would you approach it ?
Breakdown the events:
Survival: Company survives (S) or goes Bankrupt (B)
Test : Passes test § or Fails test (F)
You could then draw a tree diagram. First pair of branches showing B (prob = 0.40) and S (prob = 0.60).
From there onwards, if B occurs, you have a pair of branches with Pass § and Fail (F)
Similarly, if S occurs, you have another pair of branches with Pass § and Fail (F)
If S occurs, and then P => Prob on Pass§ branch given S occurs => Prob (Pass | Survive) = 0.85 (given)
Question: Probability of “Passing” given bankrupt ?
Since you are focused on a Probability of “Passing”, only branches BP (Bankrupt and Pass) and SP (Survive and Pass) are relevant.
Given: prob (P ) =0.55, prob (P|S) = 0.85, prob (B) =0.4 => prob (S) =0.6 [mark on tree diagram]
Mutiplying along the “Pass” branches and summing : prob (BP) + prob (SP) = Prob §
[prob (B) x prob(P | B)] + [prob (S) x prob (P|S)] = Prob §
0.40 x prob(P|B) + 0.60 x 0.85 = 0.55
prob(P|B) = 0.10