We have,
X ~ Poisson("\\lambda"), "\\lambda" > 0
For Poisson distribution mean = "\\lambda", then by the problem we have "\\lambda" = 0.4.
The p.m.f. of X is given by,
P(X = x) = "\\begin{cases}\n \\frac{e^{-0.4}.(0.4)^x}{x!} &\\text{if } x=0,1,2,...\\infty \\\\\n 0 &otherwise\n\\end{cases}"
The probability that the random variable is greater than 0
= P(X > 0)
= 1 - P(X = 0) [since the total probability is 1]
= 1 - "\\frac{e^{-0.4}.(0.4)^0}{0!}"
= 1 - 0.6703 = 0.3297 (rounded to 4 decimal places)
Answer: The probability that the random variable is greater than zero is 0.3297.
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