Suppose $300$ misprints are distributed randomly throughout a book of $500$ pages. The probability that a given page contains, at least one misprint is

1. e^{- 0.6}

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1 - e^{- 0.6}

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(0.6) e^{- 0.6}

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(0.06) e^{- 0.6}

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Solution

The correct option is D $1 - e^{- 0.6}$
Here $λ = \frac{300}{500} = 0.6$
Hence $P (X \geq 1) = 1 - P (X = 0) = 1 - \frac{e^{- 0.6} (0.6)^{0}}{0!} = 1 - e^{- 0.6}$

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Similar questions

Q. In a book of 500 pages , there are 500 misprinting errors, find the probability of at most three misprinting 3 pages selected at random from the book.

Q. In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained:

No. of misprints per page (x):	0	1	2	3	4	5
No. of pages (f):	154	95	36	9	5	1

Find the average number of misprints per page.

In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained:
No. of misprints per page (x) : 0 1 2 3 4 5
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Find the average number of misprints per page.

Q. In the first proofreading of a book containing

300

pages the following distribution of misprints was obtained:

No. of misprints per page ( $x$ )	$0$	$1$	$2$	$3$	$4$	$5$
No. of pages ( $f$ )	$154$	$95$	$36$	$9$	$5$	$1$

Find the average number of misprints per page.

From a book containing 100 pages, one page is selected randomly. The probability that the sum of the digits of the page number of the selected page is 11, is

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Suppose 300 misprints are distributed randomly throughout a book of 500 pages. The probability that a given page contains, at least one misprint is

The correct option is D 1−e−0.6Here λ=300500=0.6Hence P(X≥1)=1−P(X=0)=1−e−0.6(0.6)00!=1−e−0.6

Suppose $300$ misprints are distributed randomly throughout a book of $500$ pages. The probability that a given page contains, at least one misprint is

The correct option is D $1 - e^{- 0.6}$
Here $λ = \frac{300}{500} = 0.6$
Hence $P (X \geq 1) = 1 - P (X = 0) = 1 - \frac{e^{- 0.6} (0.6)^{0}}{0!} = 1 - e^{- 0.6}$