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) 0.4096. D) 0.7373.
finfan Wrote: ------------------------------------------------------- > 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) 0.4096. > > D) 0.7373. additon rule??? not sure P x or Y = P(x) + p(y) - P(xy) ???
D
mcf Wrote: ------------------------------------------------------- > D mcf, please explain how…
This is where I use my guess technique of picking the 3rd highest number
marlin02910 Wrote: ------------------------------------------------------- > This is where I use my guess technique of picking > the 3rd highest number LOL! i’ve done that too…
P(no default) = 0.8^5 P(1 default) = 5*0.2*0.8^4 add them up D
heha168 Wrote: ------------------------------------------------------- > P(no default) = 0.8^5 > P(1 default) = 5*0.2*0.8^4 > > add them up > D very difficult to remember these little things in the exam… I am sure to get this wrong…
It’s more complicated than that… you need to multiply by the combination. 5cbn0 = 1 5cnb1 = 5
mcf Wrote: ------------------------------------------------------- > It’s more complicated than that… you need to > multiply by the combination. > > 5cbn0 = 1 > 5cnb1 = 5 I have no clue what this is… plssss elaborate…
Probability of none defaulting is: [5!/(0!*5!)]*0.2^0*0.8^5 Probability of one defaulting: [5!/(1!*4!)]*0.2^1*0.8^4 Add them, this would give you D.
Map just gave you the full equation. You need to multiply the by combination formula. I just put the calculator key strokes (since it’s a lot easier to remember:)
probability that 0 will default is just the probability that bond 1 wont default and the probabilty that bond 2 wont default, etc. and is the same as multiplication so p(0 default) = .8*.8*.8*.8*.8 the probality that 1 bond defaults is the probability of a binomial distribution. you should look this up. it only makes since if you stare at it for a while then its easy. the example from the scheser book of beans is a good one. p(1 bond defaulting) = (5!/1! * 4!) / (.2^1*^.8^4) you add the p(none default) and p(1 defaults)
Thanks guys! Can I use the binomial distribution for most of such probability questions… i know this question is not clear enough…
Use this for any question that asks a yes/no question. But be careful regarding what the question asks. If wants to known the probability of multiple outcomes. Say, Rain on 2, 3, or 4 days in a 7 days period… you need to sum 3 separate calcs!
okie… thanks so much again…
mcf, I am assuming those are HP key strokes because I stared at my TI for a while and found nothing similar.