AP EAMCET · Maths · Probability
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
- A \(\sum_{k=3}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2\)
- B \(\sum_{k=3}^{\infty} \frac{e^{\lambda} \lambda^k}{k !}, \lambda=0.2\)
- C \(1\)- \(\sum_{k=0}^{3} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2\)
- D \(\sum_{k=0}^{3} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2\)
Answer & Solution
Correct Answer
(A) \(\sum_{k=3}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}, \lambda=0.2\)
Step-by-step Solution
Detailed explanation
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 \(\mathrm{n}=1\)…
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