Answer to Question #287853 in Statistics and Probability for Hammad

Question

A recent study found that 2 out of every 10 houses in a neighborhood have no insurance. If 5 houses are selected from 10 houses, find the probability that fewer than 4 will be insured.

Answer 1

for binomial distribution we have:

$p=0.8,n=5$

$P(x=k)=C^k_n p^k (1-p)^{n-k}$

$P(x<4)=P(0)+P(1)+P(2)+P(3)$

$P(0)=0.2^5=0.0003$

$P(1)=5\cdot0.8\cdot0.2^4=0.0064$

$P(2)=10\cdot0.8^2\cdot0.2^3=0.0512$

$P(3)=10\cdot0.8^3\cdot0.2^2=0.2048$

$P(x<4)=0.0003+0.0064+0.0512+0.2048=0.2627$