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A number of accidents that a person has in a given year is a Poisson random variable with mean$位$.

Let $X$is a number of accidents that a person has in a given year is a Poisson random variable with mean $位$.

$位=2$with $P\{位=2\}=0.6$and $位=3$

With $P\{位=3\}=0.4$

$P\{X=0\}=e^{-位}$.

Substitute

$=0.6路e^{-2}+0.4路e^{-3}$

Add the value,

$=0.10112$.

A number of $0$accidents value found to be$P\{X=0\}=0.10112$.

Exactly $3$accidents in a certain year and given that he had no accidents the preceding year

Exactly$3$accidents in a certain year

Substitute,

$=0.6\frac{2^3{e}^{-2}}{3!}+0.4\frac{3^{3e^{-3}}}{3!}$

$=0.6\frac{8e^{-2}}{6}+0.4\frac{27e^{-3}}{6}$

Simplify,

$=\frac{1}{6}(0.645+0.54)$

$=0.1979$.

$P\{X=3鈭�$he had no accidents the preceding year$\}$

$=\frac{P\{X=3,X=0\}}{P\{X=0\}}$

Substitute,

$=\frac{0.6\frac{2^3{e}^{-2}}{3!}{e}^{-2}+0.4\frac{3^{3e^{-3}}}{3!}{e}^{-3}}{0.10112}$

Divide,

$=\frac{0.019114265}{0.10112}$

$=0.189$

The exact $3$accidents in a certain year value are $P\{X=3\}=0.1979$and $P\{X=31$he had no accidents the preceding year$\}=0.189$.