# 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?

by issh 18.02.2015

Log in to add a comment

by issh 18.02.2015

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

We can assume a probability distribution for this kind of problem. We can say that the number of defective pages is a random variable X. The probability of a page being defective is random 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 E(X) = λ = mean or expected value = 5

We can assume the Poission probability distribution or the Normal probability distribution to solve this. We assume that X follows the Possion's probability distribution function.

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 probability that the number of defective pages is equal to or more than 10.

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

for info:

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

P(X=1)=0.0337 P(X=5) = 0.175

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 E(X) = λ = mean or expected value = 5

We can assume the Poission probability distribution or the Normal probability distribution to solve this. We assume that X follows the Possion's probability distribution function.

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 probability that the number of defective pages is equal to or more than 10.

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

for info:

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

P(X=1)=0.0337 P(X=5) = 0.175

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