JEE MAIN - Mathematics (2005 - No. 30)
A random variable $$X$$ has Poisson distribution with mean $$2$$.
Then $$P\left( {X > 1.5} \right)$$ equals :
Then $$P\left( {X > 1.5} \right)$$ equals :
$${2 \over {{e^2}}}$$
$$0$$
$$1 - {3 \over {{e^2}}}$$
$${3 \over {{e^2}}}$$
Explanation
In a position distribution,
$$P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}$$ ($$\lambda$$ = mean).
Now, $$P(X = r > 1.5) = P(2) + P(3) + $$ ..... $$\infty$$
$$ = 1 - \{ P(0) + P(1)\} $$
$$ = 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}}}$$
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