Answer to Question #253221 in Statistics and Probability for jil

Question

There are 1000 units of a particular product in a factory, 10 of which are defective. Assume you are a member of the quality control department tasked with determining the product's quality. So you choose three things at random from a list of 1000.
Identify which probability method is applicable to find how likely that none of them are defective and Implement it

Answer 1

The most applicable method is calculate as conditional probability likehood.

We want to find the P(A), where A is the statement: "All of the three things is non-defecrive"

This statement can be split into three:

A = B⋂C⋂D, where B - "first detail is non-defective'', C - "second detail is non-defective'', D - "third detail is non-defective''

$P(B) = {\frac {990} {1000}} ={\frac {99} {100}}$

Probability of event C provided that event B occured is:

$P(C) = {\frac {989} {999}}$

Probability of event D provided that event B and C occured is:

$P(D) = {\frac {988} {998}}$

$P(A) = P(B)*P(C)*P(D) = 0.97$ (approximately)