Question

In a city it is found that 10 accidents took place in a span of 50 days. Assuming that the number of accidents follow the Poisson distribution, the probability that there will be 3 or more accidents in a day in that city, is

• A. \( 1 - (1.02)e^{-0.2} \)
• B. \( 1 - (1.22)e^{-0.2} \)
• C. \( 1 - (1 - 0.2)e^{-0.2} \)
• D. \( 1 - \frac{1.22}{e^{0.2}} \)

Hint:

Identify the Poisson distribution parameter \( \lambda \) (average rate). Use the complement rule \( P(X \ge k) = 1 - P(X<k) \). Calculate the probabilities for \( X = 0, 1, 2 \) using the Poisson probability mass function \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \).

Answer 1

The Correct Option is B

The average number of accidents per day is \( \lambda = \frac{10}{50} = 0.
2 \).
The probability of \( k \) accidents in a day is \( P(X = k) = \frac{e^{-0.
2} (0.
2)^k}{k!} \).
We want to find \( P(X \ge 3) = 1 - P(X<3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] \).
\( P(X = 0) = \frac{e^{-0.
2} (0.
2)^0}{0!} = e^{-0.
2} \) \( P(X = 1) = \frac{e^{-0.
2} (0.
2)^1}{1!} = 0.
2 e^{-0.
2} \) \( P(X = 2) = \frac{e^{-0.
2} (0.
2)^2}{2!} = \frac{e^{-0.
2} (0.
04)}{2} = 0.
02 e^{-0.
2} \) \( P(X<3) = e^{-0.
2} + 0.
2 e^{-0.
2} + 0.
02 e^{-0.
2} = (1 + 0.
2 + 0.
02) e^{-0.
2} = 1.
22 e^{-0.
2} \) \( P(X \ge 3) = 1 - 1.
22 e^{-0.
2} = 1 - (1.
22)e^{-0.
2} \)