A Poisson distribution has a double mode at x = 1 and x = 2. The probability for x = 1 or for x = 2 of these two value is:
4e -2
A Poisson distribution has a double mode at consecutive integers \(\lambda-1\) and \(\lambda\) when \(\lambda\) is a positive integer. Modes at \(x=1, 2\) give \(\lambda = 2\).
Step 1 — Apply the Poisson PMF:
\[P(X=k)=\frac{e^{-\lambda}\lambda^{k}}{k!}\]
Step 2 — Compute each probability:
\[P(X=1)=\frac{e^{-2}\cdot 2}{1!}=2e^{-2},\quad P(X=2)=\frac{e^{-2}\cdot 4}{2!}=2e^{-2}\]
Step 3 — Add (mutually exclusive events):
\[P(X=1\text{ or }X=2)=2e^{-2}+2e^{-2}=4e^{-2}\]
Therefore the required probability is \(4e^{-2}\).
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