quality inspector

[defeated_soldier]

A machine roduces 10 units of an article in a day of which 4 are defective. The quality inspector allows releases of the products if he finds none of the 3 units chosen by him at random to be defective. What is the probabilty of quality inspector allowing the realese?



i am copying the solution given in my book...i dont understand a part of it(bolod-colored)
solution in my book:
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The total numberof ways the inspector can select 3 units = 10C3

Now,to release he must choose 3 units out of the 10-4=6 non-defective units which can occur in 6C3 ways.


so, the required probability =6C3/10C3


My question:
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please see th bold colored text........the inspector may not choose units out of 6 non-defctive units ......is not it ? THIS MAY NOT HAPPEN .....is not it ? why its then calculated ?

 

[soroban]

Hello, defeated_soldier!

A machine produces 10 units of an article in a day of which 4 are defective.
The quality inspector allows release of the products
if he finds none of the 3 units chosen by him at random to be defective.
What is the probabilty of quality inspector allowing the realese?

i am copying the solution given in my book.
i dont understand a part of it(bolod-colored)


Solution in my book:

The total number of ways the inspector can select 3 units \(\displaystyle =\:_{10}C_3\)

Now, to release he must choose 3 units out of the \(\displaystyle 10\,-\,4\:=\:6\) non-defective units
\(\displaystyle \;\;\) which can occur in \(\displaystyle _6C_3\) ways.

so, the required probability \(\displaystyle =\;\frac{_6C_3}{_{10}C_3}\;\;\) . . . all this is correct

Read the problem again . . .

In one day, the machine produces: 6 Good and 4 Defective units.

To release the production, he must select 3 of the Good units.

Got it?