Question

a survey found that customers are overcharged by price scanning systems, on average, on 1.65% of items. suppose a customer purchases 12 items. find the following probability.
a customer is overcharged on 3 items.

the probability is
(round to six decimal places as needed.)

Explanation:

Step1: Identify the binomial probability formula

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 in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Determine the values of \(n\), \(k\), and \(p\)

Here, \(n = 12\) (the number of items purchased), \(k = 3\) (the number of over - charged items), and \(p=0.0165\) (the probability of an item being over - charged)

Step3: Calculate the combination \(C(n,k)\)

$$
LATEXBLOCK0
$$

Step4: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

\(p^{k}=(0.0165)^{3}=0.0165\times0.0165\times0.0165 = 4.490625\times10^{-6}\)
\(1-p=1 - 0.0165=0.9835\)
\((1 - p)^{n - k}=(0.9835)^{9}\approx0.863797\)

Step5: Calculate the probability \(P(X = 3)\)

$$
LATEXBLOCK1
$$

Answer:

\(0.000859\)