If the average number of accidents occurring | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The number of accidents follows a Poisson distribution with parameter $\lambda = 5$.The probability mass function for a Poisson distribution is gi

Vedclass · Accepted Answer

$\frac{6}{e^5}$

Vedclass · Answer

(C) The number of accidents follows a Poisson distribution with parameter $\lambda = 5$.The probability mass function for a Poisson distribution is given by $P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$.We need to find the probability that at most one accident occurs,which is $P(X \le 1) = P(X = 0) + P(X = 1)$.For $x = 0$,$P(X = 0) = \frac{e^{-5} 5^0}{0!} = \frac{e^{-5} 	imes 1}{1} = e^{-5}$.For $x = 1$,$P(X = 1) = \frac{e^{-5} 5^1}{1!} = \frac{e^{-5} 	imes 5}{1} = 5e^{-5}$.Therefore,$P(X \le 1) = e^{-5} + 5e^{-5} = 6e^{-5} = \frac{6}{e^5}$.

$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

If the average number of accidents occurring at a particular junction on a highway in a week is $5$,then the probability that at most one accident occurs in a particular week is

Similar Questions

If two cards are drawn randomly from a pack of $52$ playing cards,then the mean of the probability distribution of the number of kings is

If the probability mass function (p.m.f.) is given by $P(X) = k \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$ and $k > 0$,and $P(X) = 0$ otherwise,then the value of $k$ is:

The probability distribution of a random variable $X$ is given by
$X = x$ $0$ $1$ $2$
$P(X = x)$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{2}{5}$

Then the variance of $X$ is

If $X$ is a Poisson variate with $P(X=0)=0.8$,then the variance of $X$ is

Suppose that a random variable $X$ follows a Poisson distribution. If $P(X=1) = P(X=2)$, then $P(X=5)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module