Mathematics, 01.07.2020 15:01 kookycookiefanx

In finance, one example of a derivative is a financial asset whose value is determined (derived) from a bundle of various assets, such as mortgages. Suppose a randomly selected mortgage in a certain bundle has a probability of 0.08 of default. (a) What is the probability that a randomly selected mortgage will not default? (b) What is the probability that nine randomly selected mortgages will not default assuming the likelihood any one mortgage being paid off is independent of the others? Note: A derivative might be an investment that only pays when all nine mortgages do not default. (c) What is the probability that the derivative from part (b) becomes worthless? That is, at least one of the mortgages defaults

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Answer from: sierracupcake0220

a) 92% probability that a randomly selected mortgage will not default

b) 47.22% probability that nine randomly selected mortgages will not default

c) 52.78% probability that the derivative from part (b) becomes worthless

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening

Suppose a randomly selected mortgage in a certain bundle has a probability of 0.08 of default.

This means that p = 0.08

(a) What is the probability that a randomly selected mortgage will not default?

Either it defaults, or it does not default. The sum of the probabilities of these outcomes is 1. So

0.08 + p = 1

p = 0.92

92% probability that a randomly selected mortgage will not default

(b) What is the probability that nine randomly selected mortgages will not default assuming the likelihood any one mortgage being paid off is independent of the others?

This is P(X = 0) when n = 9. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.08)^{0}.(0.92)^{9} = 0.4722$

47.22% probability that nine randomly selected mortgages will not default.

(c) What is the probability that the derivative from part (b) becomes worthless? That is, at least one of the mortgages defaults

Either none defect, or at least one does. The sum of the probabilities of these events is 100%. So

p + 47.22 = 100

p = 52.78

52.78% probability that the derivative from part (b) becomes worthless

Answer from: Quest

answer are

step-by-step explanation:

Answer from: Quest

not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove.

an inaccurate proof leads to errors, or stating results that are not accurate. which wil mess the probelm up you dont want this

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In finance, one example of a derivative is a financial asset whose value is determined (derived) fro...