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A book contains 100 misprints distributed randomly through out its 100 pages. The probability that a page observed at random contain atleast two mis prints is

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$\frac{1}{e}$

$1 - \frac{1}{e}$

$\frac{2}{e}$

$1 - \frac{2}{e}$

answer is D.

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Detailed Solution

$λ = \frac{100}{100} = 1$

Probability that a page contain at least two midpoints

$P (x \geq 2) = 1 - (P (x = 0) + P (x = 1))$

$= 1 - [\frac{e^{- 1}}{0!} + \frac{e^{- 1}}{1!}] = 1 - \frac{2}{e}$

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