# An insurance company found that 25% of all insurance policies

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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Ronnie Schechter
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,
$P\left(X=x\right)=n{C}_{x}{p}^{x}{q}^{n-x}$
n number o samples
x number of success
p probability of success
q probability of failure
Step 2
to find the required probability:
$P\left(X\le 8\right)=\sum _{x=0}^{8}10{C}_{x}{\left(0.75\right)}^{x}{\left(0.25\right)}^{10-x}$
$=1-\sum _{x=9}^{10}10{C}_{x}{\left(0.75\right)}^{x}{\left(0.25\right)}^{10-x}$
$=1-\left[10{C}_{9}{\left(0.75\right)}^{9}{\left(0.25\right)}^{10-9}+10{C}_{10}{\left(0.25\right)}^{10-10}\right]$
$=1-\left[0.1877+0.0563\right]$
$=1-\left[0.2440\right]$
$=0.7560$
Therefore the probability that at most eight policies are not terminated before maturity is 0.7560 or 75.60%
###### Not exactly what you’re looking for?
aquariump9
$p=0.25,q=1-p=1-0.25=0.75,n=10$
Let X be the number of policies terminated before maturity-
Probability that at most 8 policies and not terminated before maturity-
$P\left(X\le 8\right)=1-P\left(X>8\right)=1-\left[P\left(X=9\right)+P\left(X=10\right)\right]$
$=1-\left[{}^{\left\{10\right\}}{C}_{9}{\left(0.25\right)}^{9}{\left(0.75\right)}^{1}{+}^{10}{C}_{10}{\left(0.25\right)}^{10}\right]$
PSK=1-(0.0000286+0.000009536)
$=1-0.00002955=0.999997$
###### Not exactly what you’re looking for?
nick1337

An insurance company found that 25% of all insurance policies are terminated before their maturity date
$⇒$ Probability  of  policy termination $p=\frac{25}{100}=\frac{1}{4}$
Probability of policy not terminating  $=1-1/4=3/4$
15 policies are randomly selected $⇒n=15$
$p\left(x\right){=}^{n}{C}_{x}{p}^{x}{q}^{n-x}$
probability that more than 8 but less than 11 policies are terminated before maturing
$⇒x=9,10$
$⇒probability=p\left(9\right)+p\left(10\right)$
${=}^{15}{C}_{9}\left(\frac{1}{4}{\right)}^{9}\left(\frac{3}{4}{\right)}^{6}{+}^{15}{C}_{10}\left(\frac{1}{4}{\right)}^{10}\left(\frac{3}{4}{\right)}^{5}$
$=\left({3}^{5}/{4}^{15}\right)\left(5005\cdot 3+3003\right)$
$=\left({3}^{5}/{4}^{15}\right)\left(18018\right)$
=0.0041
= 0.41% is the probability that more than 8 but less than 11 policies are terminated before maturing