Answer to Question #291713 in Statistics and Probability for Shaaz

Question

1.   Each sample of water has a 10% chance of containing a particular organic pollutant. Assume that the samples are independent with regard to the presence of the pollutant.

Let X =the number of samples that contain the pollutant in the next 18 samples analyzed. Then X is a binomial random variable with p = 0:10 and   n = 18.

 

i.  Find the probability that in the next 18 samples, exactly 2 contain the pollutant.

ii.  Find the probability that 3≤ X ≤ 5

iii. Find the probability that X ≥ 2

Answer 1

n=18

p=0.1

q=1-0.1=0.9

$Using~\binom{n}{x}p^xq^{n-x}\\ i)p(x=2)=\binom{18}{2}(0.1)^2(0.9)^{18-2}\\ =\binom{18}{2}(0.1)^2(0.9)^{16}\\ =153 \times 0.01 \times 0.1853=0.8438\\ ii)p(3 \leq x \leq 5)=\binom{18}{3}(0.1)^3(0.9)^{18-3}+\binom{18}{4}(0.1)^4(0.9)^{18-4}+\binom{18}{5}(0.1)^5(0.9)^{18-5}\\ =\binom{18}{3}(0.1)^3(0.9)^{15}+\binom{18}{4}(0.1)^4(0.9)^{14}+\binom{18}{5}(0.1)^5(0.9)^{13}\\ =0.1680+0.0700+0.0218\\ =0.2598\\ iii)p(x \geq 2)=1-p(x < 2)\\ =1-[\binom{18}{0}(0.1)^0(0.9)^{18}+\binom{18}{1}(0.1)^1(0.9)^{17} ]\\ =1-[0.1501+0.3002]\\ =1-0.4503\\ =0.5497.$