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kseniaru

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tT
发表于 2013-4-17 18:11 | 只看该作者

Probabilities Question

For a certain class of junk bonds, the probability of default in a given year is 20%. The default prbability of bonds are independent of each other. Portfolio of 5 bonds, what is the probability that 0 (zero) and 1 (one) bond among the portfolio of 5 (five) bonds defaults?
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ogoluwa

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2#
发表于 2013-4-17 18:11 | 只看该作者
I think the question is asking about the probability that any one bond defaults, not a specific one.
Prob for 0 bonds to default = (.8)^5 = .32768
Prob for 1 bond to default = 1C5*(.2)(.8)^4 = .4096

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Matori

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3#
发表于 2013-4-17 18:11 | 只看该作者
Yes, cfagoal2, you are right.
I didn’t comprehend the question correctly. I took that as probability of Bond0 and Bond1 defaulting and the rest not defaulting.

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Colum

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4#
发表于 2013-4-17 18:12 | 只看该作者
thanks.
what are the original formulas for these in the text?
Can you explain this one: 1C5*(.2)(.8)^4 = .4096

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needhelp1700

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5#
发表于 2013-4-17 18:12 | 只看该作者
It’s just the multiplication rule.
There is a .8 probability that each bond will NOT default. Therefore, for all 5 bonds to NOT default it’s .8 x .8 x .8 x .8 x .8
For 1 bond to default, it’s the probability of 1 bond defaulting (.2) multiplied by the probability of the other 4 bonds NOT defaulting (.8)
Therefore it’s .2 x .8 x .8 x .8 x .8

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Carson

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6#
发表于 2013-4-17 18:12 | 只看该作者
don’t think that’s it…cfagoal2 had the right answer

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andytrader

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7#
发表于 2013-4-17 18:12 | 只看该作者
from cfagoal2’s answer:
it is Probability of 1 bond defaulting * number of ways it could be that one from the pack of 5 * probability of rest 4 not defaulting.
.2 * 5C1 * .8 * .8 * .8 * .8

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soddy1979

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8#
发表于 2013-4-17 18:12 | 只看该作者
any 1 of the five bonds can default. How many ways can that happen? You choose 1 bond to default out of 5. and that answer is 5C1 (combinations) = 5 ways.

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Viceroy

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9#
发表于 2013-4-17 18:12 | 只看该作者
The question is asking for the probability of default
default probability =p =0.2
no default probability =q =1p =0.8
0 bond default probability = (.2)^5(.8)^0=0.0003
1 bond default probability = (.2)^4(.8)^1=0.0013
correct me if my understanding is not right

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xilinx_altera

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10#
发表于 2013-4-17 18:12 | 只看该作者
A bond either defaults, or does not. You use the binomial probability distribution.
0 Default = 5C0 *[(1.2)^(50)]*[(1.8)^(55)] = 1*[(.8)^5]*1 = .32768
1 Default = 5C1 *(1.2)^(51)]*[(1.8)^(54)] = 5 *[(.8)^4]*(.2) = .4096
nCr is the combination function on the calculator. This equations gives “the probability of exactly x success in n trials”.

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