CA Foundation · Quantitative Aptitude · Theoretical Distributions
The number of accidents per week at a busy crossing in Nagpur follows a Poisson distribution with mean 2. Given e^-2 = 0.1353, what is the probability that exactly 3 accidents occur in a given week?
The probability is 0.1804. Using the Poisson formula, P(X=3) = e^-2 × 2^3 divided by 3 factorial, which is 0.1353 × 8 / 6 = 0.1804. The mean of 2 is the parameter, and the required count is 3 accidents in the week.
- A0.1804Correct
- B0.2707
- C0.0902
- D0.5413
Explanation
P(X=3) = e^-2 × 2^3 / 3! = 0.1353 × 8 / 6 = 0.1804. Option B is P(X=2) or P(X=1), which uses the wrong x value (0.1353×2). Option C uses 2^3/3!... mistakenly dividing by 12 instead of 6.
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