Schweser Qbank #1523 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) .5904 b) .0819 c) .4096 d) .7373 Ans: D The outcome follows a binomial distribution where n = 5 and p = 0.2. In this case p(0) = 0.8^5 = 0.3277 and p(1) = 5 × 0.8^4 × 0.2 = 0.4096, so P(X=0 or X=1) = 0.3277 + 0.4096. My question is this … for p(1) = 5 × 0.8^4 × 0.2, why do you multiply by 5?

That’s the formula: n!/k!*(n-k)! = 5!/4!*1! gives you 5

Conceptually there are 5 ways in which 1 bond can default. If it were two, it would be 10 (5!/3!2!) ways in which two bonds can default.

I get it. So diagramming this, it would look like this: D = Default N = No Default DNNNN NDNNN NNDNN NNNDN NNNND So that’s why you multiply by 5 right? And in the case of p(0) = .8^5, you “multiply by 1” because there is only one way for all to default.

remembering the formula map1 stated probably takes less time than diagramming it out, but it is useful to diagram to kind of figure out what the forumula is doing.

topher Wrote: ------------------------------------------------------- > Schweser Qbank #1523 > > 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) .5904 > b) .0819 > c) .4096 > d) .7373 > > Ans: D > The outcome follows a binomial distribution where > n = 5 and p = 0.2. In this case p(0) = 0.8^5 = > 0.3277 and p(1) = 5 × 0.8^4 × 0.2 = 0.4096, so > P(X=0 or X=1) = 0.3277 + 0.4096. > > My question is this … for p(1) = 5 × 0.8^4 × > 0.2, why do you multiply by 5? This is kinda like a GMAT problem. Answer is Probability that none will default + 1 will default. .8^5 + .8^4*.2 now we have 5! ways to arrange these w/ no constraints. but we will have 4 non defaults and 1 defualt. (4 ways to arrange the non defaults) so really its 5C1 --> 5!/4! --> so .8^5+ 5(.8^4*.2) --> .32768 + .4096 =.73728 ~.7373 D

haha this is where I got this from for GMATClub

topher Wrote: ------------------------------------------------------- > Schweser Qbank #1523 > > 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) .5904 > b) .0819 > c) .4096 > d) .7373 > > Ans: D > The outcome follows a binomial distribution where > n = 5 and p = 0.2. In this case p(0) = 0.8^5 = > 0.3277 and p(1) = 5 × 0.8^4 × 0.2 = 0.4096, so > P(X=0 or X=1) = 0.3277 + 0.4096. > > My question is this … for p(1) = 5 × 0.8^4 × > 0.2, why do you multiply by 5? Should I also reply ?? well let me try see question is in two part assume P(0)= x ; probablity that 0 no will default ; case 1 also P(1)= y ; probablity that 1 will default .; case 2 here its not possible that case 1 and case 2 will happen simultanously. P(0 or 1 ) = P(0) + P(1) - P(0 and 1) here P(0 and 1 )= 0. P®= n©r . p^r . (1-p)^n-r here p = 0.2 , probability of default in a given year is 0.2 r = 0 . multiplying 5 is because of term n©r term in bionomial probablity n=5, r=1 so 5©1= no of ways of choosing one default among five.

Three parts to this problem: 1. Zero default: .8 x.8 x.8 x.8 x.8 =.32768 2. One default: .2x.8x.8x.8x.8 =.08192. But this is only one possible combination in which exactly one bond could default. There are a total of 5 different ways that exactly one bond could default, thus you need to multiply .08192 x 5 = .4096. You could also use the combo formula here to determine the different number of ways this that exactly one bond could default {5!/[(5-1)!1!]=5}. 3. Add the probabilities together, since you are calculating the probability of one or the other occuring: .32768 + .4096 = .73728. Answer: D