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  • The mean and variance of a Poisson distribution are 2 and 1 respectively. Find the probability that there will be at least two successes in 5 trials.Option: 1</s

The mean and variance of a Poisson distribution are 2 and 1 respectively. Find the probability that there will be at least two successes in 5 trials.

Option: 1

0.301


Option: 2

0.408


Option: 3

0.556


Option: 4

0.624


Answers (1)

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To find the probability that there will be at least two successes in 5 trials in a Poisson distribution with mean 2, we can use the cumulative probability function of the Poisson distribution.
Let X be the random variable representing the number of successes in 5 trials. The Poisson distribution with mean \lambda  (lambda) is given by:
\mathrm{P(X=k)=\left(e^{\wedge}(-\lambda)^* \lambda^{\wedge} k\right) / k ! }
Where, \mathrm{k}=number of successes

To find the probability of at least two successes ( 2 or more) in 5 trials, we need to sum the probabilities of getting 2,3,4, or 5 successes.

P( at least two successes) \mathrm{ =P(X=2)+P(X=3)+ \\ P(X=4)+P(X=5) }

- Exactly two successes:\mathrm{ (1)^{\wedge} 2^* e^{\wedge}(-1)^* 5 ! / 2 ! 3 != 0.3678794411714423}
- Exactly three successes: \mathrm{(1)^{\wedge} 3^* e^{\wedge}(-1)^* 5 ! / 3 ! 2 ! = 0.1461637995910642}
- Exactly four successes:\mathrm{ (1)^{\wedge} 4^* \theta^{\wedge}(-1)^* 5 ! / 4 ! 1 ! = 0.03347053250968108}
- Exactly five successes:\mathrm{ (1)^{\wedge} 5^{\star} e^{\wedge}(-1)=0.007965625}


The sum of these probabilities is:

\mathrm{0.367+0.1461+0.033+0.007=0.556 }
So, the probability that there will be at least two successes in 5 trials in the given Poisson distribution is approximately \mathrm{ 0.556 , or 56 \%. }

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SANGALDEEP SINGH

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