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In a city, 10 accidents take place in a span of 50 days.Assuming that the number of accidents follow the Poisson distribution,the probability that three or more accidents occur in a day, is
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Verified Answer
The correct answer is:
$\sum_{k=3}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2$
For poisson distribution,
$P(\mathrm{X}=k)=\frac{e^{-\lambda} \lambda^k}{k !}$
Where $\lambda=$ mean of distribution $=n \rho$
$k=\text { probability of success }$
Here, 10 accidents take place in 50 days.
So, $\mathrm{p}=\frac{10}{50}=\frac{1}{5}$ and $\mathrm{n}=1$
$\therefore \lambda=1 \times \frac{1}{5}=0.2$
Probability that three or more accidents occur in a day,
$\begin{aligned}
& P(x \geq 3)=P(x=3)=P(x=4)+\ldots . \\
& =\sum_{k=3}^\infty \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2
\end{aligned}$
$P(\mathrm{X}=k)=\frac{e^{-\lambda} \lambda^k}{k !}$
Where $\lambda=$ mean of distribution $=n \rho$
$k=\text { probability of success }$
Here, 10 accidents take place in 50 days.
So, $\mathrm{p}=\frac{10}{50}=\frac{1}{5}$ and $\mathrm{n}=1$
$\therefore \lambda=1 \times \frac{1}{5}=0.2$
Probability that three or more accidents occur in a day,
$\begin{aligned}
& P(x \geq 3)=P(x=3)=P(x=4)+\ldots . \\
& =\sum_{k=3}^\infty \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2
\end{aligned}$
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