Miller, Chapter 2: Probabilities Question

[lRRAngle]

Hi David,

Would you please explain how your calculation of 2 x 30% x (1-30%)=42% for the question in bold? Specifically why are you multiplying 3% times 2 and no simply 30% times 70%?

You own two bonds. Both bonds have a 30% probability of defaulting. Their default probabilities
are statistically independent. What is the probability that both bonds default? What is the
probability that only one bond defaults? What is the probability that neither bond defaults?

Thanks

 

[David Harper CFA FRM]

HI @lRRAngle Say these are called Bond A and Bond B. There can four outcomes if we care about the sequence:

• Bond A survives and Bond B survives; probability = 70%*70% = 49.0%
• Bond A survives and Bond B defaults; probability = 70%*30% = 21.0%
• Bond A defaults and Bond B survives; probability = 30%*70% = 21.0%
  
• Bond A defaults and Bond B defaults; probability = 30%*30% = 9.0%; Total = 49.0% + 21.0% + 21.0% + 9.0% = 100.0%

There are four outcomes but only three events if we do not care about sequence (or maybe I should say, if we do not care about the identity of the defaulted bond): 1. Neither defaults; 2. One defaults, 3. Both default. This is why the binomial probability of one default contains the binomial coefficient. The Prob(x = 1 default | p = 30%, n = 2) = 30%^1*(1-30%)^(2-1)*C(2,1) = 30%*70%*2, where C(2,1) is the number of different ways we can have one default among two bond. The answer is 2.

So, here as elsewhere, the wording of the question matters greatly. These questions (writing by S&W) are:

What is the probability that both bonds default? What is the probability that only one bond defaults? What is the probability that neither bond defaults?

... In regard to the question "What is the probability that only one bond defaults?", notice how that could be either Bond A or Bond B defaulting, which is two different outcomes (but only a single event, if we define an event according to the number of defaults). I hope that's helpful!