CA Foundation · Quantitative Aptitude · Theoretical Distributions
The number of accidents per month at a busy junction in Chennai follows a Poisson distribution with mean 2. Using e^-2 = 0.1353, what is the probability that there is no accident in a given month?
The probability of no accident is e raised to minus 2, which equals 0.1353. With mean 2, P(X=0) = e^-2 * 2^0 / 0!, and the power and factorial terms both equal one, leaving only e^-2.
- A0.2706
- B0.1353Correct
- C0.8647
- D0.5413
Explanation
P(X=0) = e^-m * m^0 / 0! = e^-2 = 0.1353. The value 0.8647 is 1 - 0.1353, the probability of at least one accident, so it answers a different question. 0.2706 is P(X=1) = 2 * 0.1353.
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