Question

1. suppose that 80% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. consider groups of 17 trucks and let x be the number of trucks in a group that have passed the inspection. what is the probability that there will be between 10 and 12 trucks (inclusive) which pass the inspection? a) 0.80 b) 0.0680 c) 0.2308 d) 0.5120 e) 0.2041

Explanation:

Step1: Identify the distribution

This is a binomial - distribution problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 17$, $p=0.8$, and we want to find $P(10\leq X\leq12)=P(X = 10)+P(X = 11)+P(X = 12)$.

Step2: Calculate $C(n,k)$ for $k = 10$

$C(17,10)=\frac{17!}{10!(17 - 10)!}=\frac{17!}{10!7!}=\frac{17\times16\times15\times14\times13\times12\times11}{7\times6\times5\times4\times3\times2\times1}=19448$.
$P(X = 10)=C(17,10)\times(0.8)^{10}\times(0.2)^{7}=19448\times0.1073741824\times0.0000128 = 0.0267$.

Step3: Calculate $C(n,k)$ for $k = 11$

$C(17,11)=\frac{17!}{11!(17 - 11)!}=\frac{17!}{11!6!}=\frac{17\times16\times15\times14\times13\times12}{6\times5\times4\times3\times2\times1}=12376$.
$P(X = 11)=C(17,11)\times(0.8)^{11}\times(0.2)^{6}=12376\times0.08589934592\times0.000064 = 0.0680$.

Step4: Calculate $C(n,k)$ for $k = 12$

$C(17,12)=\frac{17!}{12!(17 - 12)!}=\frac{17!}{12!5!}=\frac{17\times16\times15\times14\times13}{5\times4\times3\times2\times1}=6188$.
$P(X = 12)=C(17,12)\times(0.8)^{12}\times(0.2)^{5}=6188\times0.06871947674\times0.00032 = 0.1370$.

Step5: Calculate the total probability

$P(10\leq X\leq12)=P(X = 10)+P(X = 11)+P(X = 12)=0.0267 + 0.0680+0.1370 = 0.2308$.

Answer:

C. 0.2308