Lesson Accepting/rejecting shipments via acceptance procedures

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Accepting/rejecting shipments via acceptance procedures


Problem 1

A company receives a shipment of  16  items.  A random sample of  4  items is selected,
and the shipment is rejected if any two of these items proves to be defective.
What is the probability of accepting a shipment containing  4  defective items ?

Solution

The probability that the shipment will be  highlight%28highlight%28accepted%29%29  is equal to the probability to find 0 or 1 
defective items in the random sub-set of 4 (four) items in the set of 16 items.


The number of all possible quadruples of 16 items is  C%5B16%5D%5E4 = %2816%2A15%2A14%2A13%29%2F%281%2A2%2A3%2A4%29 = 1820.

The number of all possible quadruples of the form (0 defective,4 good) is  C%5B4%5D%5E0%2AC%5B12%5D%5E4 = 1*495 = 495. 

The number of all possible quadruples of the form (1 defective,3 good) is  C%5B4%5D%5E1%2AC%5B12%5D%5E3 = 4*220 = 880.

So the probability that the shipment will be accepted is

    P = %28495+%2B+880%29%2F1820 = 1375%2F1820 = 275%2F364 = 0.755494505 = 75.55%  (rounded).    ANSWER

Problem 2

A shipment of  10  items has two defective and eight non defective units.
In the inspection of the shipment,  a sample of  4  (four)  units will be selected and tested.
If a defective unit is found,  the shipment of  10  units will be rejected.
If a random sample of four items is tested,  what is the probability that the shipment will be rejected?

Solution

The probability that the shipment will be rejected is equal to the probability to find at least
one defective item in the random set of 4 (four) items.


The number of all possible quadruples of 10 items is  C%5B10%5D%5E4 = %2810%2A9%2A8%2A7%29%2F%281%2A2%2A3%2A4%29 = 210.


The number of all possible quadruples of the form (1 defective,3 good) is  C%5B2%5D%5E1%2AC%5B8%5D%5E3 = 2%2A%28%288%2A7%2A6%29%2F%281%2A2%2A3%29%29 = 112.


The number of all possible quadruples of the form (2 defective,2 good) is  C%5B2%5D%5E2%2AC%5B8%5D%5E2 = 1%2A%28%288%2A7%29%2F%281%2A2%29%29 = 28.


So the probability that the shipment will be rejected is

    P = %28112+%2B+28%29%2F210 = 140%2F210 = 2%2F3.    ANSWER


My other lessons on  Probability  in this section are
    - Probabilistic analysis of a court verdicts
    - A company bids on two separate contracts
    - Advanced probability problems related to combinations
    - Using cumulative sums and relevant standard functions to solve problems on Binomial Distributions
    - Miscellaneous binomial distribution problems
    - Binomial distribution problems on overbooking flights
    - Advanced probability problems on binomial distribution
    - Probabilistic analysis of testing procedures in health care
    - Probabilistic analysis of testing procedures in industry
    - Probability problems on games
    - Probability problem on winning a many-rounds game
    - A Math circle level probability problem on a lottery game
    - Upper league entertainment probability problems
    - Upper league probability problems on Stars and Bars methodology
    - Upper league problem to maximize winning in a game with 20-sided rolling die
    - Upper league problems on conditional probability
    - Upper League geometric probability problems
    - Probability problems on long chains of related events
    - Probability problems similar to a coinciding birthdays problem
    - Using empirical rules to determine normal distribution probabilities
    - OVERVIEW of lessons on Probability, section 3

Use this file/link  OVERVIEW of lessons on Probability  to navigate over all my lessons on  Probability problems  (section 1)  in this site.

Use this file/link  OVERVIEW of my additional lessons on Probability  to navigate over all my lessons on additional  Probability problems  (section 2)  in this site.

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.



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