What is the probability that, on a given day, Rooneys will receive at most two web-based enquiries for holiday packages?

Steps to Solve

1. Identify the Poisson Distribution Parameters

In this problem, the average number of enquiries per day ($\lambda$) is 5. We need to calculate the probability of receiving at most 2 enquiries, which means we are looking for $P(X \leq 2)$.

2. Use the Poisson Probability Formula

The probability of receiving $k$ enquiries is given by the formula:

$$ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} $$

Here, we will find $P(X = 0)$, $P(X = 1)$, and $P(X = 2)$.

3. Calculate Each Probability

   • For $k = 0$: $$ P(X = 0) = \frac{e^{-5} 5^0}{0!} = e^{-5} $$

   • For $k = 1$: $$ P(X = 1) = \frac{e^{-5} 5^1}{1!} = 5e^{-5} $$

   • For $k = 2$: $$ P(X = 2) = \frac{e^{-5} 5^2}{2!} = \frac{25e^{-5}}{2} $$

4. Sum the Probabilities

Now we find the total probability of receiving at most 2 enquiries:

$$ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) $$ Substituting the results from above:

$$ P(X \leq 2) = e^{-5} + 5e^{-5} + \frac{25e^{-5}}{2} $$

5. Calculate the Final Probability

Factor out $e^{-5}$:

$$ P(X \leq 2) = e^{-5} \left(1 + 5 + \frac{25}{2}\right) $$ $$ P(X \leq 2) = e^{-5} \left(\frac{2 + 10 + 25}{2}\right) = e^{-5} \frac{37}{2} $$

6. Substituting Value of $e^{-5}$ and Calculate

Using the approximate value of $e^{-5} \approx 0.00674$:

$$ P(X \leq 2) \approx 0.00674 \cdot \frac{37}{2} \approx 0.1247 $$