Question

Ms. Bergen is a loan officer at Coast Bank and Trust. From her years of experience, she estimates that the probability is .025 that an applicant will not be able to repay his or her installment loan. Last month she made 40 loans. Use the poisson approximation to the binomial.
a. What is the probability that three loans will be defaulted?
b. What is the probability that at least 3 loans will be defaulted?

Answer 1

Answer:

a) 0.0613 b)0.0803

Step-by-step explanation:

Ms. Bergen estimates that the probability is 0.025 that an applicant will not be able to repay his or her installment loan.

p = 0.025

Let's consider that an applicant is not be able to repay  his or her installment loan as a ''success''

p (success) = 0.025

Last month she made 40 loans ⇒ n = 40

For the poisson approximation to the binomial we need to calculate n.p that  will be the λ parameter in our poisson approximation

$n.p=40.(0.025)=1$

λ=n.p=1

Let's rename λ = j

In our poisson approximation :

$f(k,j)=\frac{e^{-j} .j^{k} }{k!}$

f(k,j) is the probability function for our poisson variable where we calculated j,e is the euler number and k is the number of success :

$f(k,1)=\frac{e^{-1} .1^{k} }{k!}$

For a) We are looking the probability of 3 success :

$f(3,1)=\frac{e^{-1} .1^{3} }{3!}=0.0613$

For b) We are looking for the probability of at least 3 success

If ''L'' is the number of success

$P(L\geq 3)=1-P(L\leq 2)$

$P(L\leq 2)=P(L=0)+P(L=1)+P(L=2)$

$P(L\leq 2)=f(0,1)+f(1,1)+f(2,1)$

$P(L\leq 2)=e^{-1} +e^{-1}+\frac{e^{-1}}{2} =e^{-1}(1+1+\frac{1}{2} )$

$P(L\geq 3)=1-P(L\leq 2)=1-e^{-1}(1+1+\frac{1}{2} )=0.0803$

Answer 2

The probability that three loans will default is 0.0613

The probability that at least 3 loans will default is 0.0803

Calculations and Parameters:

Ms. Bergen estimates that the probability is 0.025 that an applicant will not be able to repay his or her installment loan.

p = 0.025

Hence, we consider that an applicant is not able to repay his or her installment loan as a ''success''

p (success) = 0.025

Last month she made 40 loans ⇒

n = 40

For the Poisson approximation to the binomial, we need to calculate n.p which will be the λ parameter in our Poisson approximation

n.p= 40.(0.025) =1

λ=n.p=1

Let's rename λ = j

In our Poisson approximation :

f(k,1) = e^-j.j^k/k!

Hence, the probability of 3 success is:

f(3,1)= e^-1.1^3/3!

=0.0613.

The probability of at least 3 successes is:

If  ''L'' is the number of successes.

P(L≥ 3) = 1- P(L≤ 2)

P(L≥ 3)= 0.0803.

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