Answer to Question #199928 in Statistics and Probability for kotjinjo

Question

An insurance company found that 25% of all insurance policies are terminated before their maturity date. Assume that 15 polices are randomly selected from the company’s policy database. Required: Out of the 15 randomly selected policies; a) What is the expected number of policies to be terminated before maturing? (2) b) What is the probability that no policy will be terminated before maturity? (2) c) What is the probability that all policies will be terminated before maturing? (2) d) What is the probability that at least two policies will be terminated? (4) e) What is the probability that more than 8 but less than 11 policies will be terminated? (5) f) What is the probability that eight policies will not be terminated? (3) g) What is the probability that 18 policies will be terminated?

Answer 1

Let $X=$ the number of policies which will be terminated: $X\sim Bin(n, p).$

Given $n=15, p=0.25$

a)

$$E(X)=np=15(0.25)=3.75$$

b)

$$P(X=0)=\dbinom{15}{0}0.25^0(1-0.25)^{15-0}$$

$$\approx0.013363$$

c)

$$P(X=15)=\dbinom{15}{15}0.25^{15}(1-0.25)^{15-15}$$

$$\approx0.000000001$$

d)

$$P(X\geq2)=1-P(X=0)-P(X=1)$$

$$1-\dbinom{15}{0}0.25^{0}(1-0.25)^{15-0}$$

$$-\dbinom{15}{1}0.25^{1}(1-0.25)^{15-1}$$

$$\approx0.919819$$

e)

$$P(8<X<11)=P(X=9)+P(X=10)$$

$$=\dbinom{15}{9}0.25^{9}(1-0.25)^{15-9}$$

$$+\dbinom{15}{10}0.25^{10}(1-0.25)^{15-10}$$

$$\approx 0.0033981+0.0006796$$

$$\approx0.004078$$

f)

$$P(X\not=8)=1-P(X=8)$$

$$=1-\dbinom{15}{8}0.25^{8}(1-0.25)^{15-8}$$

$$\approx 0.986893$$

g)

$$P(X=18)=0$$