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If the number of accidents at a busy intersection follows a Poisson distribution with a mean of 3.6 per week, what is the probability that exactly 2 accidents will occur in a given period of time?

A) 0.491
B) 0.110
C) 0.268
D) 0.183
Please select the correct option.

1 Answer

6 votes

Final answer:

The probability of exactly 2 accidents occurring at a busy intersection with a mean of 3.6 accidents per week, according to the Poisson distribution, is approximately 17.67%, which rounds to option D) 0.183.

Step-by-step explanation:

The probability that exactly 2 accidents will occur in a busy intersection that has accidents following a Poisson distribution with a mean (λ) of 3.6 per week can be calculated using the Poisson probability formula:

P(X=k) = (λ^k * e^-λ) / k!

Where k is the exact number of accidents we want to find the probability for, in this case, k=2.

P(X=2) = (3.6^2 * e^-3.6) / 2! = (12.96 * 0.027323) / 2 = 0.35345488 / 2 = 0.17672744.

Therefore, the probability that exactly two accidents will occur is approximately 0.1767 or 17.67%, which corresponds to the option D) 0.183 after rounding to three decimal places.

User Nikita Chernov
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