Question

A cab hire firm has two cabs, which it hires out day by day. The number of demands for cabs on each day follows a Poisson distribution with mean of 1.5. The probability of days on which neither cab is used is (Use \( e^{-1.5} = 0.2231 \)):

• A. 0.1353
• B. 0.2231
• C. 0.7231
• D. 0.018

Hint:

The Poisson distribution is frequently used for modeling the number of arrivals or demands in a fixed interval of time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A Poisson distribution is given by \( P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \), where \( \lambda = 1.5 \). "Neither cab is used" means the demand \( X = 0 \).

Step 2: Key Formula or Approach:
Substitute \( \lambda = 1.5 \) and \( k = 0 \) into the formula.

Step 3: Detailed Explanation:
\[ P(X=0) = \frac{1.5^0 \cdot e^{-1.5}}{0!} \]
Since \( 1.5^0 = 1 \) and \( 0! = 1 \):
\[ P(X=0) = e^{-1.5} \]
Given \( e^{-1.5} = 0.2231 \).

Step 4: Final Answer:
The probability is 0.2231.