Question

22% of all tires manufactured by a company are defective. if 20 different tires are randomly selected from this company what is the probability that exactly 9 tires are not defective?
a. 0.0484
b. 0.0010
c. 0.0132
d. 0.9946

Explanation:

Step1: Determine probabilities of defective and non - defective

The probability of a tire being defective $p = 0.22$, so the probability of a tire being non - defective $q=1 - p=1 - 0.22 = 0.78$.

Step2: Identify the binomial formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{n - k}\times q^{k}$, where $n$ is the number of trials, $k$ is the number of successes, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 20$, $k = 9$.

Step3: Calculate the combination $C(20,9)$

$C(20,9)=\frac{20!}{9!(20 - 9)!}=\frac{20!}{9!11!}=\frac{20\times19\times\cdots\times12}{9!}=167960$.

Step4: Calculate the probability

$P(X = 9)=C(20,9)\times(0.22)^{20 - 9}\times(0.78)^{9}$
$P(X = 9)=167960\times(0.22)^{11}\times(0.78)^{9}$
$(0.22)^{11}\approx3.13\times10^{-8}$, $(0.78)^{9}\approx0.075$
$P(X = 9)=167960\times3.13\times10^{-8}\times0.075\approx0.0010$

Answer:

B. 0.0010