A firm hires out 3 cars every day. The demand for a car on each day is distributed as a poisson distribution with mean 2.0. The probability that some demand is refused on a day is `(e^(-2)=0.135)`

Options

A. 0.2177
B. 0.1912
C. 0.1455
D. 0.0192

Answer

To solve the problem, we need to find the probability that some demand for cars is refused on a given day. This occurs when the demand exceeds the number of cars available for hire, which is 3 in this case. We can model the demand using a Poisson distribution with a mean (λ) of 2.0.

### Step-by-step Solution:

1. **Define the Random Variable**:
Let \( X \) be the number of cars hired per day. Since the demand follows a Poisson distribution with a mean of \( \lambda = 2.0 \), we can express this as:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

2. **Identify the Event of Interest**:
We need to find the probability that some demand is refused. This happens when the demand \( X \) exceeds the number of cars available, which is 3. Therefore, we need to calculate:
\[ P(X > 3) = 1 - P(X \leq 3) \]

3. **Calculate \( P(X \leq 3) \)**:
We can express \( P(X \leq 3) \) as:
\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]
Using the Poisson formula, we calculate each term:
- For \( k = 0 \):
\[ P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} \cdot 1 = e^{-2} \]
- For \( k = 1 \):
\[ P(X = 1) = \frac{e^{-2} \cdot 2^1}{1!} = e^{-2} \cdot 2 \]
- For \( k = 2 \):
\[ P(X = 2) = \frac{e^{-2} \cdot 2^2}{2!} = e^{-2} \cdot \frac{4}{2} = 2e^{-2} \]
- For \( k = 3 \):
\[ P(X = 3) = \frac{e^{-2} \cdot 2^3}{3!} = e^{-2} \cdot \frac{8}{6} = \frac{4}{3} e^{-2} \]

4. **Sum the Probabilities**:
Now we sum these probabilities:
\[ P(X \leq 3) = e^{-2} + 2e^{-2} + 2e^{-2} + \frac{4}{3}e^{-2} \]
Combine the terms:
\[ P(X \leq 3) = e^{-2} \left( 1 + 2 + 2 + \frac{4}{3} \right) = e^{-2} \left( 5 + \frac{4}{3} \right) \]
Convert \( 5 \) to a fraction:
\[ 5 = \frac{15}{3} \Rightarrow P(X \leq 3) = e^{-2} \left( \frac{15 + 4}{3} \right) = e^{-2} \cdot \frac{19}{3} \]

5. **Calculate \( P(X > 3) \)**:
Now we can find \( P(X > 3) \):
\[ P(X > 3) = 1 - P(X \leq 3) = 1 - e^{-2} \cdot \frac{19}{3} \]

6. **Substitute the Value of \( e^{-2} \)**:
Given that \( e^{-2} = 0.135 \):
\[ P(X > 3) = 1 - 0.135 \cdot \frac{19}{3} \]
Calculate \( 0.135 \cdot \frac{19}{3} \):
\[ = 0.135 \cdot 6.3333 \approx 0.855 \]
Therefore:
\[ P(X > 3) = 1 - 0.855 = 0.145 \]

### Final Answer:
The probability that some demand is refused on a day is approximately \( 0.145 \).