a survey found that customers are overcharged by price scanning systems, on average, on 2.15% of items. suppose a customer purchases 12 items. find the following probability.
a customer is overcharged on 2 items.
the probability is
(round to six decimal places as needed.)

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Accepted Answer

# 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 on 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), $k = 2$ (the number of over - charged items), and $p=0.0215$ (the probability of an item being over - charged)

## Step3: Calculate the combination $C(n,k)$
$$
\begin{align*}
C(12,2)&=\frac{12!}{2!(12 - 2)!}\
&=\frac{12!}{2!×10!}\
&=\frac{12\times11\times10!}{2\times1\times10!}\
& = 66
\end{align*}
$$

## Step4: Calculate $p^{k}$ and $(1 - p)^{n - k}$
$p^{k}=(0.0215)^{2}=0.00046225$
$1-p = 1-0.0215 = 0.9785$
$(1 - p)^{n - k}=(0.9785)^{10}\approx0.807777$

## Step5: Calculate the probability $P(X = 2)$
$$
\begin{align*}
P(X = 2)&=C(12,2)\times p^{2}\times(1 - p)^{10}\
&=66\times0.00046225\times0.807777\
&\approx66\times0.0003731\
&\approx0.024625
\end{align*}
$$

# Answer:
$0.024625$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the binomial probability formula

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

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

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

Step5: Calculate the probability \(P(X = 2)\)

Answer: