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Sum Rule & Bayes’ Theorem

  • The unconditional probability of event B is equal to the sum of joint probabilities of event (A,B) and the probability of event (A’,B). Here A’ is the probability of not happening of A
    • The joint probability of events A and B is the product of conditional probability of B, given A has occurred and the unconditional probability of event A

    • We know that P(AB) = P(B/A) * P(A)
    • Also P(BA)= P(A/B) *  P(B)
    • Now equating both P(AB) and P(BA) we get:   

 

P(B) can be further broken down using sum rule defined above: 
 

Example

Out of a group of 100 patients being treated for chronic back trouble, 25% are chosen at random to receive a new, experimental treatment as opposed to the more usual muscle relaxant-based therapy which the remaining patients receive. Preliminary studies suggest that the probability of a cure with the standard treatment is 0.3, while the probability of a cure from the new treatment
is 0.6.

1. How many patients (on an average) out of the 100 patients selected at random would be cured?

  1. 30
  2. 40
  3. 37.5
  4. 42.5

 
2. Some time later, one of the patients returns to thank the staff for her complete recovery.
What is the probability that she was given the new treatment?

  1. 0.375
  2. 0.425
  3. 0.4
  4. 0.425
     
Solution

1. C.
25% are given new treatment =>75% are given old treatment.
P(Cure) = P(Cure/New) * P(New) + P(Cure/Old) * P(Old) = 0.375
So out of 100 patients 37.5 will get cured
 
2. C.
Apply Bayes’ Theorem
P (New / Cure) = P (Cure / New) * P (New) / P (Cure) = 0.6 * 0.25 / 0.375 = 0.40
 


 

Example

Calculate the probability of a subsidiary and parent company both defaulting over the next year. Assume that the subsidiary will default if the parent defaults, but the parent will not necessarily default if the subsidiary defaults. Assume that the parent had a 1 year PD = 0.5% and the subsidiary has 1 year PD of 0.9%.

  1. 0.45%
  2. 0.5%
  3. 0.545%
  4. 0.55%
     
Solution

B.
P (S| P) = 1 = P(P & S)/P(P)
P (P & S) = P(P) = 0.5%
 

 





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