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Books front a certain publisher contain an average of 1 misprint per page. What is the probability that on at least one page?

Sagot :

The probability for at least 1 page is 0.667 .

Probability is the department of arithmetic concerning numerical descriptions of the way likely an event is to occur, or how probable it is that a proposition is actual. The probability of an event is a number between 0 and 1, where, roughly speaking, zero shows the impossibility of the event, and 1 suggests truth.

Possibility = the wide variety of methods of achieving success. The whole wide variety of possible outcomes. As an example, the opportunity of flipping a coin and its being heads is ½ because there's 1 way of having a head and the total quantity of viable outcomes is 2 (a head or tail). We write P(heads) = ½.

The form opportunity is from old French probability (14 c.) and at once from Latin probabilities (nominative probabilitas) "credibility, possibility," from probabilistic (see likely). The mathematical sense of the time period is from 1718.

As there's an average of 1 misprint in line with the web page, the possibility of a minimum of 5 misprints is 1 - P(0, 1, 2, 3, or 4 misprints)

P(0) = e-1/0!

P(1) = e-1/1!

P(2) = e -1/2!

P(3) = e-1/3!

P(4) = e-1/4!  

P(0, 1, 2, 3, or 4 misprints) = e-1(1 + 1 + half of + 1/6 + 1/24) = 2 17/24 e-1 = .99634

Then, P(5 or more misprints) is 1 - .99634 = .00366

Then, the P(at the least 1 page in a 300-page ebook has at least 5 misprints) = 1 - P(0 pages have at least 5 misprints).

P(0 pages have at the least 5 misprints) = P(300 of 300 pages have 0, 1, 2, 3, or 4 misprints) =.99634 300

(you could view this as C(300, 300) p300 in case you desire) = 0.332881690573629

Then, P(1 or greater pages have at least 5 misprints) = 1 - 0.332881690573629 = 0.667118309426371

Learn more  about probability here https://brainly.com/question/24756209

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