Question

Based on past experience, 1 % of the telephone bills mailed to house-holds in Hong Kong are incorrect. If a sample of 10 bills is selected, find the probability that at least one bill will be incorrect. Do this using two probability distributions (the binomial and the Poisson) and briefly compare and explain your results.

Answer 1

Answer with explanation:

Given : The probability that telephone bills mailed to house-holds in Hong Kong are incorrect.=0.01

Binomial distribution :-

$P(x)=^nC_xp^x(1-p)^{n-x}$

If a sample of 10 bills is selected, then the probability that at least one bill will be incorrect :-

$P(x\geq1)=1-P(0)\\\\=1-^{10}C_{0}(0.1)^0(0.9)^{10}\\\\=1-(0.9)^{10}=0.6513$

Hence, the probability that at least one bill will be incorrect =0.6513

Poisson distribution:

$P(x;\mu)=\dfrac{e^{-\mu}\mu^x}{x!}$

Mean : $\mu=np=10\times0.1=1$

Then , If a sample of 10 bills is selected, then the probability that at least one bill will be incorrect :-

$P(x\geq1)=1-P(0)\\\\=1-\dfrac{e^{-1}1^0}{0!}\\\\=1-0.3678=0.6321$

Hence, the probability that at least one bill will be incorrect =0.6321