Question

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 at least 10 but fewer than 13 trucks pass the inspection

Answer 1

Answer:

0.2308 = 23.08% probability that at least 10 but fewer than 13 trucks pass the inspection

Step-by-step explanation:

For each truck, there are only two possible outcomes. Either they pass the inspection, or they do not. The probability of a truck passing the inspection is independent of other trucks. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

80% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection.

This means that $p = 0.8$

17 trucks:

This means that $n = 17$

What is the probability that at least 10 but fewer than 13 trucks pass the inspection

$P(10 \leq X < 13) = P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{17,10}.(0.8)^{10}.(0.2)^{7} = 0.0267$

$P(X = 11) = C_{17,11}.(0.8)^{11}.(0.2)^{6} = 0.0680$

$P(X = 12) = C_{17,12}.(0.8)^{12}.(0.2)^{5} = 0.1361$

$P(10 \leq X < 13) = P(X = 10) + P(X = 11) + P(X = 12) = 0.0267 + 0.0680 + 0.1361 = 0.2308$

0.2308 = 23.08% probability that at least 10 but fewer than 13 trucks pass the inspection