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is known that two defective copies of a commercial software program were erroneously sent to a shipping lot that has now a total of 90 copies of the program. A sample of copies will be selected from the lot without replacement. Round your answers to five decimal places (e.g. 98.76543).

(a) If three copies of the software are inspected, determine the probability that exactly one of the defective copies will be found.
(b) If three copies of the software are inspected, determine the probability that both defective copies will be found.
(c) If 88 copies are inspected, determine the probability that both copies will be found. (Hint: Work with the copies that remain in the lot.)

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umairfeb221999

Answer:

a) The probability of finding exactly 1 defective copy out of 3 is 0.06517.

b) The probability of finding both defective copies out of 3 is 0.00075.

c) The probability of finding both defective copies among 88 inspected copies is 0.95580.

Step-by-step explanation:

a) Defective copies can be first, second, or third in 3 inspected copies, thus

           [tex]P( 1st\ one ) =(\frac{2}{90} )* (\frac{88}{89})*(\frac{87}{88})[/tex]

          [tex]P( 2nd\ one ) = (\frac{88}{90})*(\frac{2}{89} )*(\frac{87}{88})[/tex]

          [tex]P( 2nd\ one ) = (\frac{88}{90})*(\frac{87}{89})*(\frac{2}{88} )[/tex]

So,

           [tex]P(exactly \ 1 \ out\ of\ 3)=3*\frac{2*88*87}{88*89*90} =0.06517[/tex]

b) Similarly,

          [tex]P(Exactly \ 2\ out\ of\ 3)=3*\frac{1*2*88}{88*89*90} =0.00075[/tex]

c) Now, Choosing 88 copies, it is same as choosing 2 copies which would not be inspected so, the probability of detecting the 2 defected copies in the 88 inspected copies while leaving 2 copies (not inspected) is:

            [tex]P(both\ get\ found)=\frac{88}{90} * \frac{87}{89} =0.95580[/tex]

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