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If a random variable X follows poisson's distribution such that P(X=2)=9P(X=4)+90P(X=6)P(X=2) = 9P(X=4) + 90P(X=6), then the mean of X is

Solution

✅ Correct Option: 1

Using P(X=r)=e−λλrr!P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!} and cancelling e−λλ2e^{-\lambda}\lambda^2:

12=9λ224+90λ4720\dfrac12=\dfrac{9\lambda^2}{24}+\dfrac{90\lambda^4}{720}, i.e. λ4+3λ2−4=0\lambda^4+3\lambda^2-4=0.

So λ2=1\lambda^2=1 and λ=1\lambda=1; the mean is 1.

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