Question

In a big city, $5$ accidents take place over a period of $100$ days. If the numebr of accidents follows P.D., the probability that there will be $2$ accidents in a day is

• A. $\frac{e^{-5}{5}^2}{2!}$
• B. $\frac{e^{-05}{5}^2}{2!}$
• C. $\frac{e^{-005}(0.05{)}^2}{2!}$
• D. $\frac{e^5{5}^2}{2!}$

Answer 1

Answer: (C)

$\frac{e^{-005}(0.05{)}^2}{2!}$

By using Poisson distribution,

$P(x;\mu )=\frac{e^{-\mu }{\mu }^x}{x!}$

$\mu :$ The mean number of successes that occur in a specified region.

$x:$ The actual number of successes that occur in a specified region.

$P(x;\mu ):$ The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is

$\mu =\frac{5}{100}=0.05$

$x=2$

$P(2;005)=\frac{e^{-0.05}{0.05}^2}{2!}$