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Q.If the variance of a Poisson distribution is 2, then P(X=2)P(X = 2) is : (A) 4e24e^2 (B) 2e22e^2 (C) 2e2\dfrac{2}{e^2} (D) 4e2\dfrac{4}{e^2}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★
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Poisson variance =λ=2=\lambda=2, so P(X=2)=e−2 222!=2e2P(X=2)=\dfrac{e^{-2}\,2^2}{2!}=\dfrac{2}{e^2}.

Poisson: P(X=x)=e−λλxx!P(X=x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}, where both mean and variance equal λ\lambda.

  1. Variance =λ=2=\lambda = 2 (mean and variance coincide for Poisson).
  2. Substitute x=2x=2: P(X=2)=e−2⋅222!P(X=2) = \dfrac{e^{-2}\cdot 2^2}{2!}. …

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