An insurance company found that 25% of all insurance policies are terminated before their maturity date. Assume that 10 polices are randomly selected from the company’s policy database. Assume a Binomial experiment. Required: What is the probability that at most eight policies are not terminated before maturity?

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An insurance company found that 25% of all insurance policies are terminated before their maturity date. Assume that 10 polices are randomly selected from the company’s policy database. Assume a Binomial experiment.  Required:  What is the probability that at most eight policies are not terminated before maturity?

Ronnie Schechter · Accepted Answer

Step 1  Given,  An insurance company found that all insurance policies terminated before their maturity date.  The Probability of insurance terminated before the maturity date is 25%  q=0.25  Then, the Probability of insurance not terminated before the maturity date is, p=1−0.25=0.75  Assume that n=10 policies are randomly selected from the company’s policy database.   Let success be the insurance not terminated before the maturity date the required calculation can be defined by the formula,  P(X=x)=nCxpxqn−x  n number o samples  x number of success  p probability of success  q probability of failure  Step 2  to find the required probability:  P(X≤8)=∑x=0810Cx(0.75)x(0.25)10−x  =1−∑x=91010Cx(0.75)x(0.25)10−x  =1−[10C9(0.75)9(0.25)10−9+10C10(0.25)10−10]  =1−[0.1877+0.0563]  =1−[0.2440]  =0.7560  Therefore the probability that at most eight policies are not terminated before maturity is 0.7560 or 75.60%

aquariump9 · Accepted Answer

p=0.25,q=1−p=1−0.25=0.75,n=10&amp;nbsp;Let X be the number of policies terminated before maturity-&amp;nbsp;Probability that at most 8 policies and not terminated before maturity-&amp;nbsp;P(X≤8)=1−P(X&amp;gt;8)=1−[P(X=9)+P(X=10)]&amp;nbsp;=1−[&amp;nbsp;{10}C9(0.25)9(0.75)1+10C10(0.25)10]&amp;nbsp;=1-(0.0000286+0.000009536)&amp;nbsp;=1−0.00002955=0.999997

nick1337 · Accepted Answer

An insurance company found that 25% of all insurance policies are terminated before their maturity date ⇒ Probability  of  policy termination p=25100=14 Probability of policy not terminating  =1−1/4=3/4 15 policies are randomly selected ⇒n=15 p(x)=nCxpxqn−x probability that more than 8 but less than 11 policies are terminated before maturing ⇒x=9,10 ⇒probability=p(9)+p(10) =15C9(14)9(34)6+15C10(14)10(34)5 =(35/415)(5005⋅3+3003) =(35/415)(18018) =0.0041  = 0.41% is the probability that more than 8 but less than 11 policies are terminated before maturing

An insurance company found that 25% of all insurance policies

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