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# The number of accidents in a year involving taxi drivers in a city follows a poisson distribution with mean equal to $3$. Out of $1000$ taxi drivers find approximately the number of driver with no accident in a year .$[e^{-3} = 0.0498].$

Toolbox:
• A random variable $X$ is said to have a poisson distribution if the probability mass function of $X$ is
• $P(X=x)=\large\frac{e^{\Large -\lambda }\lambda^x}{x!}$$\qquad (x=0,1,2........$ for some $\lambda > 0)$
• Constants of a poisson distribution :
• Mean=Variance=$\lambda$
• The parameter of the Poisson distribution is $\lambda$
• A Poisson random variable corresponds to rare events.
Step 1:
Let $X$ be the random variable denoting the number of accidents involving taxi drivers in a year.
$X\sim P(3)$
$P(X=x)=\large\frac{e{-3}3^x}{x!}$
Step 2:
Probability that a taxi driver is not involved in any accident =$P(X=0)=e^{-3}=0.0498$
Out of 1000 drivers the expected number of drivers who will not be involved in any accident =$n\times probability$
$\Rightarrow 1000\times 0.0498=50$(approx)