A local factory uses a manufacturing process in which 70% of the final products meet quality standards and 30% are found to be defective. If

Hmm.  

Suppose there are 1,000,000 items, 700,000 of which are good, 300,000 of which are defective.

The probability that the first one you choose is good is 700,000/1,000,000 = 0.7. 

You now have 699,999 good ones left out of a total of 999,999, so the probability of selecting a second good one is 699,999/999,999 ≈ 0.6999997

You now have 699,998 good ones left out of a total of 999,998, so the probability of selecting a third good one is 699,998/999,998 ≈ 0.6999994

The overall probability is therefore 0.7*0.6999997*0.6999994 ≈ 0.343 ≈ 0.7³

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Odds of picking the first one are 0.7. Now there are 69 left out of 70 for the second choice,so odds of picking second good one are 0.69.Similarly for the third choice,the odds are 0.68. 

With each choice,multiply with odds of picking the next,and the result is 0.7x0.69x0.68=32.844% likelihood that they are all good.

Mathcad, what makes you think that there were 70 to start with?

correction, prob is 70/100 x 69/99 x 68/98. Approx 33.85%.

If the probability that an object is acceptable is 0.70, and it is assumed that the probability of whether or not an object is acceptable or defective is independent from whether or not other objects are wither acceptable or defiectime(that is, whether or not an object is defective does not influence whether or not any other objects are defective), I believe that the best answer is:

0.70 x 0.70 x 0.70  =  0.343  34.3%

If there were only 100 objects to choose from and 70 are know to be accepable, then 70/100 x 69/99 x 68/98 would be the correct way to proceed; however, if this is a production line that may create thousands every hour, then 0.70³ is a better estimate.

Yes, that is what I would go for too Geno :)

Confusion seems to be creeping in here.(and apologies for previous careless answer!)---The 70%(and the 30%) is INDEPENDANT of the number of the factory's products. So no matter what the number of products,we are always dealing with the ratios 70/100 and 30/100.Using this scale to keep the problem as simple as possible,this boils down to 

First choice. Probability of picking a good one= 70/100 ( no matter what the number of products!) Now,keeping the scale as simple as possible,

Second choice,now have 99 choices left with 69 good ones.Prob= 69/99

Third choice follows as                                                        68/98

so prob is 70/100x69/99x68/98,regardless of the number of products. Note that I am NOT assuming that there are only 100 products,but using the percentages to keep the problem simple.You will find,and I am not prepared to do it here,that if you start with any number of products and scale down to percentages,you will get the same answer.

Mathcad, you ARE assuming that there are 100 products.

 Alan has tried to show you where you logic falls in a heap