Question

The number of persons joining a cinema ticket counter in a minute follows a Poisson distribution with parameter 6, then the probability that at least one and at most five persons join the queue in a particular minute is

• A. $e^{-6} \times 6 (25.48)$
• B. $e^{-6} \left( \frac{6}{2} + \frac{6^3}{3!} + \frac{6^4}{4!} \right)$
• C. $6 \times e^{-6} (29.8)$
• D. $e^{-6} \left(6 + \frac{6^2}{2} + \frac{6^3}{3!} + \frac{6^4}{4!} \right)$

Hint:

For Poisson distribution problems involving a range of values, calculate the probability for each value in the range and sum them up.

Answer 1

The Correct Option is C

Step 1: Identify the parameter of the Poisson distribution.
The parameter is $ \lambda = 6 $.
Step 2: Define the event of interest.
We want to find $ P(1 \le X \le 5) $, where $ X $ is the number of persons joining the queue.
Step 3: Recall the PMF of a Poisson distribution.
$ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} $
Step 4: Express the desired probability.
$ P(1 \le X \le 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) $
Step 5: Calculate each individual probability.
$ P(X=1) = \frac{e^{-6} 6^1}{1!} = 6e^{-6} $
$ P(X=2) = \frac{e^{-6} 6^2}{2!} = 18e^{-6} $
$ P(X=3) = \frac{e^{-6} 6^3}{3!} = 36e^{-6} $
$ P(X=4) = \frac{e^{-6} 6^4}{4!} = 54e^{-6} $
$ P(X=5) = \frac{e^{-6} 6^5}{5!} = 64.8e^{-6} $
Step 6: Sum the probabilities.
$ P(1 \le X \le 5) = 6e^{-6} + 18e^{-6} + 36e^{-6} + 54e^{-6} + 64.8e^{-6} = 178.8e^{-6} $
Step 7: Match with the options.
$ 178.8e^{-6} = 6 \times 29.8 \times e^{-6} = 6 \times e^{-6} (29.8) $
Thus, the probability is $ \boxed{6 \times e^{-6} (29.8)} $.