CA Foundation · Quantitative Aptitude · Theoretical Distributions

A Poisson distribution has P(X = 1) = P(X = 2). What is the probability P(X = 0) for this distribution?

Equating e^-m·m with e^-m·m²/2 gives m = 2. Therefore P(X = 0) = e^-m = e^-2. The mean of the distribution is 2.

Explanation

P(1) = e^-m m and P(2) = e^-m m²/2. Setting them equal gives m = m²/2, so m = 2. Then P(0) = e^-2. A common error is to take m = 1, which gives e^-1.

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