At a local post office, on average , 3 customers are in line when the post office closes each day. The probability P, that exactly n customers are in line when the post office closes can be modeled by the equation `P = (3^(n)e^(-3))/(n!)`. Given that `e^(-3) = 0.05`. Which of the following values is closest to the probability that exactly 2 customers are in line when the post office closes?

Options

A. 0.08
B. 0.11
C. 0.15
D. 0.23

Answer

To find the probability that exactly 2 customers are in line when the post office closes, we can use the given probability formula:

\[ P(n) = \frac{3^n e^{-3}}{n!} \]

where \( n \) is the number of customers in line. We are given that \( e^{-3} = 0.05 \) and we need to calculate \( P(2) \).

### Step 1: Substitute \( n = 2 \) into the formula

\[ P(2) = \frac{3^2 e^{-3}}{2!} \]

### Step 2: Calculate \( 3^2 \)

\[ 3^2 = 9 \]

### Step 3: Substitute \( e^{-3} \)

Since \( e^{-3} = 0.05 \), we can substitute this value into the equation:

\[ P(2) = \frac{9 \cdot 0.05}{2!} \]

### Step 4: Calculate \( 2! \)

\[ 2! = 2 \times 1 = 2 \]

### Step 5: Substitute \( 2! \) into the equation

Now we have:

\[ P(2) = \frac{9 \cdot 0.05}{2} \]

### Step 6: Calculate the numerator

\[ 9 \cdot 0.05 = 0.45 \]

### Step 7: Divide by \( 2 \)

\[ P(2) = \frac{0.45}{2} = 0.225 \]

### Step 8: Approximate the result

The value \( 0.225 \) is approximately \( 0.23 \).

Thus, the probability that exactly 2 customers are in line when the post office closes is approximately:

\[ \boxed{0.23} \]