 # Simple binomial problem

For a certain class of junk bonds, the probability of default in a given year is 0.2. Whether one bond defaults is independent of whether another bond defaults. For a portfolio of five of these junk bonds, what is the probability that zero or one bond of the five defaults in the year ahead? A) 0.5904. B) 0.0819. C) 0.7373. D) 0.4096. Your answer: D was incorrect. The correct answer was C) 0.7373. The outcome follows a binomial distribution where n=5 and p=0.2. In this case p(0)=0.85 = 0.3277 and p(1)=5 x 0.84 x 0.2=0.4096, so P(X=0 or X=1)= 0.3277 + 0.4096. Why do you include N in the calculation of P(1) and not P(0)?? Thanks

let’s use D for a bond defaulting and N for a bond not defaulting for no bonds defaulting in the five years ahead, there’s only one combination/pattern: DDDDD = (.8)(.8)(.8)(.8)(.8) = (.8^5) for 1 bond defaulting in the five years ahead, there’s 5 combinations/patterns: NDDDD or DNDDD or DDNDD or DDDND or DDDDN = (.8)(.8)(.8)(.8)(.2) = (.8^4)(.2) x 5

alternatively, you could use the binomial formula: ( n! / (n - r)! ) * (percentage default)^® * (percentage non-default)^(1-r) for zero bonds defaulting: ( 5! / (5 - 0)! ) * (0.2^0) * (.8^5) = 0.3277 for 1 bond defaulting: ( 5! / (5 - 1)! ) * (0.2^1) * (.8^4) = 0.4096 0.3277 + 0.4096 = C) 0.7373

Ahh yes or the good old nCr button 5 C 1 = 5 I just need to remember combinations when dealing with binomial distributions. I would have had to calculate the amount of cobinations for the first one as well had it had any number of bonds that defaulted too correct?