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:a) 0.0613 b)0.0803Step-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 = 0.025Let's consider that an applicant is not be able to repay  his or her installment loan as a ''success''p (success) = 0.025Last month she made 40 loans ⇒ n = 40For the poisson approximation to the binomial we need to calculate n.p that  will be the λ parameter in our poisson approximation[tex]n.p=40.(0.025)=1[/tex]λ=n.p=1Let's rename λ = jIn our poisson approximation :[tex]f(k,j)=\frac{e^{-j} .j^{k} }{k!}[/tex]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 :[tex]f(k,1)=\frac{e^{-1} .1^{k} }{k!}[/tex]For a) We are looking the probability of 3 success :[tex]f(3,1)=\frac{e^{-1} .1^{3} }{3!}=0.0613[/tex]For b) We are looking for the probability of at least 3 successIf ''L'' is the number of success[tex]P(L\geq 3)=1-P(L\leq 2)[/tex][tex]P(L\leq 2)=P(L=0)+P(L=1)+P(L=2)[/tex][tex]P(L\leq 2)=f(0,1)+f(1,1)+f(2,1)[/tex][tex]P(L\leq 2)=e^{-1} +e^{-1}+\frac{e^{-1}}{2} =e^{-1}(1+1+\frac{1}{2} )[/tex][tex]P(L\geq 3)=1-P(L\leq 2)=1-e^{-1}(1+1+\frac{1}{2} )=0.0803[/tex]