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The Number of accidents in a year attributed to taxi driver in a city follows poison distribution with mean 3.out of 1000 taxi drivers ,find approximately the number of drivers with

1)no accident in a year

2)more than 3 accidents in a year.

($e^{-1}=0.3679,e^{-2} = 0.1353,e^{-3}= 0.0498$

1 Answer
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For poison distribution $P(x=r) = \frac{ e^{-m}.m^{r}}{r!}$ where x =0,1,2

Given m=3

i) No.accidents in a year, i.e.$r=0$

$P(x=0) = \frac {e^{-3} \times (3)^0 }{0!}$

$P(x=0)=\frac {0.0498\times 1}{1}$

$P(x=0)=0.0498$

therefore expected number of drivers with no accidents

=$N\times P(x=0)$

=$1000\times 0.0498$

=$49.8 \approx 50$

ii)More than 3 accidents in a year

$P(more \ than \ 3 \ accident) = 1-p[x=0,1,2]$

=$1-[\frac{e^{-3}.3^{0}}{0!}+\frac{e^{-3}.3^1}{1!}+\frac{e^{-3}.(3)^{2}}{2!}]$

= $1-[0.0498+0.1494+0.2241]$

=$1-0.4233$

$P(more \ than \ 3 \ accident)=0.5767$

Expected number of drivers with more then 3 accidents

= $N\times p(more \ than \ 3 \ accidents)$

=$1000 (0.5767)$

$=576.7 \approx =577$

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