A manufactured product on an average has $2$ defects per unit of product produced. If the number of defects follows P.D., the probability of finding zero defects is

e^{- 2}

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1 - e^{- 2}

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\frac{e^{- 2} 2^{1}}{∠ 1}

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e^{- 002}

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Solution

The correct option is A $e^{- 2}$
By using Poisson distribution,
$P (x; μ) = \frac{e^{- μ} μ^{x}}{x!}$
: The mean number of successes that occur in a specified region.
x: The actual number of successes that occur in a specified region.
P(x; ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is
$μ = 2$ (defects per unit of product produced)
x = 0 (zero defects)
$P (0; 2) = \frac{e^{- 2} 2^{0}}{0!} = e^{- 2}$

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A manufactured product on an average has 2 defects per unit of product produced. If the number of defects follows P.D., the probability of finding zero defects is

A manufactured product on an average has $2$ defects per unit of product produced. If the number of defects follows P.D., the probability of finding zero defects is