SOLUTION: Suppose that a box of ten items from a manufacturing company is known to contain two defective and eight nodefective items. A sample of three items is selected at random. Find the

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Question 1157532: Suppose that a box of ten items from a manufacturing company is known to contain two defective and eight nodefective items. A sample of three items is selected at random. Find the probability the sample contains one defective and two nondefective items.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(54018) About Me  (Show Source):
You can put this solution on YOUR website!
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Suppose that a box of ten items from a manufacturing company is known to contain two defective and eight non-defective
items. A sample of three items is selected at random. Find the probability that the sample contains one defective
and two non-defective items.
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To solve this problem, it is convenient to consider ordered triples - samples of three items, where their order 
does matter.


In this problem, the number of all such possible triples is  10*9*8.


The number of all possible ordered triples consisting of one defective and two non-defective items is (2*8*7)*3.


Here the product 8*7 represents the number of all ordered pairs of non-defective items;
the factor 2 represents the choice of one from the two defective items, 
and factor 3 represents three possible positions for one defective item in the ordered triple.


The probability is the ratio of the number of favorable triples to the total number of triples

    P = %28%282%2A8%2A7%29%2A3%29%2F%2810%2A9%2A8%29 = %282%2A7%2A3%29%2F%2810%2A9%29 = %282%2A7%29%2F%2810%2A3%29 = 7%2F%285%2A3%29 = 7%2F15.


ANSWER.  The probability is 7%2F15.

Solved.



Answer by greenestamps(13382) About Me  (Show Source):
You can put this solution on YOUR website!


The total number of possible ways of choosing 3 of the 10 items is, literally, "10 choose 3", which is

C%2810%2C3%29=%2810%2A9%2A8%29%2F%283%2A2%2A1%29=120

The "good" outcomes are the ones in which we choose 1 of the 2 defective items and 2 of the 8 non-defective items:

C%282%2C1%29%2AC%288%2C2%29=%282%29%28%288%2A7%29%2F%282%2A1%29%29=2%2A28=56

By the basic definition of probability, the probability of choosing one defective and two non-defective items is then

56%2F120=7%2F15

ANSWER: 7/15