Probability
Poisson Distribution
Grade 12
Question:
<p>At a telephone enquiry system, the number of phone calls regarding relevant enquiry follow Poisson's distribution with an average of 5 phone calls during 10 min time interval. The probability that there is atmost one phone call during a 10 min time period, is</p>
<p>(a) \(\frac{6}{5e}\)</p>
<p>(b) \(\frac{6}{e^5}\)</p>
<p>(c) \(\frac{6}{55}\)</p>
<p>(d) \(\frac{6}{5e^5}\)</p>
Step-by-Step Solution
Key Concept: Apply the Poisson probability formula: $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$. Sum the probabilities for $k=0$ and $k=1$.
<p>For a Poisson distribution with parameter $\lambda = 5$, the probability mass function is $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$. The probability of at most one call is $P(X \leq 1) = P(X=0) + P(X=1) = \frac{e^{-5} \cdot 5^0}{0!} + \frac{e^{-5} \cdot 5^1}{1!} = e^{-5} + 5e^{-5} = 6e^{-5} = \frac{6}{e^5}$.</p>
Correct Answer: B