In a city, 10 accidents take place in a span of 50 days.Assuming that the number of accidents follow the…

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
  1. $\sum_{k=3}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2$
  2. $\sum_{k=3}^{\infty} \frac{e^{\lambda} \lambda^k}{k !}, \lambda=0.2$
  3. $1$- $\sum_{k=0}^{3} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2$
  4. $\sum_{k=0}^{3} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2$

Solution

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, $\begin{aligned} & P(x \geq 3)=P(x=3)=P(x=4)+\ldots . \\ & =\sum_{k=3}^\infty \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2 \end{aligned}$

Asked in: AP EAMCET 2016

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