a survey finds customers are overcharged by price scanning systems, on average, on 2.20% of items. suppose a customer purchases 18 items. find the following probability.
the customer is overcharged on at most 2 items.
the probability that the customer is overcharged on at most 2 items is
(round to four decimal places as needed.)

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Accepted Answer

# Explanation:
## Step1: Identify the binomial distribution parameters
The binomial distribution formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 18$ (number of trials, i.e., number of items), $p=0.022$ (probability of over - charging an item), and $C(n,k)=\frac{n!}{k!(n - k)!}$. We want to find $P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)$

## Step2: Calculate $P(X = 0)$
$$
\begin{align*}
C(18,0)&=\frac{18!}{0!(18-0)!}=1\
P(X = 0)&=C(18,0)\times(0.022)^{0}\times(1 - 0.022)^{18-0}\
&=1\times1\times(0.978)^{18}\
\end{align*}
$$
Using a calculator, $(0.978)^{18}\approx0.6774$

## Step3: Calculate $P(X = 1)$
$$
\begin{align*}
C(18,1)&=\frac{18!}{1!(18 - 1)!}=\frac{18!}{1!17!}=18\
P(X = 1)&=C(18,1)\times(0.022)^{1}\times(1 - 0.022)^{18-1}\
&=18\times0.022\times(0.978)^{17}\
\end{align*}
$$
$(0.978)^{17}\approx0.6927$, $18\times0.022\times0.6927 = 18\times0.01524 = 0.2743$

## Step4: Calculate $P(X = 2)$
$$
\begin{align*}
C(18,2)&=\frac{18!}{2!(18-2)!}=\frac{18\times17\times16!}{2\times1\times16!}=153\
P(X = 2)&=C(18,2)\times(0.022)^{2}\times(1 - 0.022)^{18 - 2}\
&=153\times(0.022)^{2}\times(0.978)^{16}\
\end{align*}
$$
$(0.022)^{2}=0.000484$, $(0.978)^{16}\approx0.7083$, $153\times0.000484\times0.7083\approx153\times0.000343\approx0.0525$

## Step5: Sum up the probabilities
$P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\approx0.6774+0.2743 + 0.0525=0.9942$

# Answer:
$0.9942$

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Question

Explanation:

Step1: Identify the binomial distribution parameters

Step2: Calculate \(P(X = 0)\)

Step3: Calculate \(P(X = 1)\)

Step4: Calculate \(P(X = 2)\)

Step5: Sum up the probabilities

Answer: