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There are 1000 pages in a book out of which 100 pages are defective. What is the probability

that out of first 50 pages 10 pages will be defective?

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that out of first 50 pages 10 pages will be defective?

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We assume a
known probability distribution for this kind of problem. The population is big enough (100 pages). We can say that the
number of defective pages is a random variable X. The probability of any
randomly selected page being defective is equal and = 100/1000 = 0.1

In the first 50 pages, the expected number of defective pages is 50 * 0.1 = 5. Let X denote the number of defective pages in the first 50 pages.

We can have 0 defective pages, 1 or 2 or 5 or 10 or 19 or 20 or up to 50 defective pages in the first 50 pages of the book. That means,

0 <= X <= 50 and we have that E(X) = λ = mean value or expected value = 5

We can assume the Poission probability distribution or the Normal probability distribution to solve this. We don’t know the standard deviation for the normal distribution. We assume that X follows the Poisson's probability distribution function. It is a discrete probability distribution.

Let us calculate the probabilities, with λ = 5.

P(X=10) = 0.0181 = the probability that the number of defective pages is 10.

P(X >= 10) <= 0.145. This is the upper limit on the probability that the number of defective pages is equal to or more than 10.

P(X <= 10) >= 0.855, so probability that the number of defective pages is less than or equal to10, is 0.855

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for info:

we can also calculate P(X=0) = 0.006738

P(X=1)=0.03369 P(X=2) = 0.0842 P(X=3) = 0.1404

P(X=4) = 0.1755 P(X=5) = 0.1755 P(X=6) = 0.1462

P(X=7) = 0.1044 P(X=8) = 0.0653 P(X=9) = 0.0362

In the first 50 pages, the expected number of defective pages is 50 * 0.1 = 5. Let X denote the number of defective pages in the first 50 pages.

We can have 0 defective pages, 1 or 2 or 5 or 10 or 19 or 20 or up to 50 defective pages in the first 50 pages of the book. That means,

0 <= X <= 50 and we have that E(X) = λ = mean value or expected value = 5

We can assume the Poission probability distribution or the Normal probability distribution to solve this. We don’t know the standard deviation for the normal distribution. We assume that X follows the Poisson's probability distribution function. It is a discrete probability distribution.

Let us calculate the probabilities, with λ = 5.

P(X=10) = 0.0181 = the probability that the number of defective pages is 10.

P(X >= 10) <= 0.145. This is the upper limit on the probability that the number of defective pages is equal to or more than 10.

P(X <= 10) >= 0.855, so probability that the number of defective pages is less than or equal to10, is 0.855

=================================

for info:

we can also calculate P(X=0) = 0.006738

P(X=1)=0.03369 P(X=2) = 0.0842 P(X=3) = 0.1404

P(X=4) = 0.1755 P(X=5) = 0.1755 P(X=6) = 0.1462

P(X=7) = 0.1044 P(X=8) = 0.0653 P(X=9) = 0.0362