A random variable X follow Poisson distribution such that P(X=1) = 2P…

The correct answer is Option (2) → \(\displaystyle\frac{1}{e}\)

For a Poisson distribution with mean \(\lambda\),

    \( \displaystyle P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!} \)

The given condition becomes

    \( \displaystyle \lambda e^{-\lambda} = 2 \times \frac{\lambda^2 e^{-\lambda}}{2} \)

    \(\lambda = \lambda^2\)

Since \(\lambda \gt 0\), \(\lambda = 1\). Then

    \( \displaystyle P(X = 0) = e^{-1} = \frac{1}{e} \)