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a bag contains articles of 4 different kinds-periodical,novel,newspaper,hardcover.WHAT...

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kausik911 | Student, Undergraduate | eNoter

Posted February 16, 2012 at 2:46 PM via web

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a bag contains articles of 4 different kinds-periodical,novel,newspaper,hardcover.WHAT IS THE SMALLEST NUMBER OF ARTICLES IN THE BAG SATISFYING FOLLOWING CONDITIONS

when 4 articles are drawn from the bag without replacement the following events are equally likely-

1. the selection of 4 periodicals.

2. the selection of 1 novel and 3 periodicles

3. the selection of 1 news paper.1 novel & 2 periodicles

4. the selection of 1 article of each kind

WHAT IS THE SMALLEST NUMBER OF ARTICLES IN THE BAG SATISFYING THESE CONDITIONS?HOW MANY OF THESE ARE OF EACH KIND?

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michaelpaulheart | eNoter

Posted February 16, 2012 at 5:29 PM (Answer #1)

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The smallest number of articles is 16, 4 of each kind.

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mubashirak | Student, College Freshman | eNotes Newbie

Posted February 17, 2012 at 3:54 AM (Answer #2)

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can you please explain

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etotheeyepi | Student, Undergraduate | Valedictorian

Posted February 17, 2012 at 4:20 AM (Answer #3)

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Yes, an explanation would be the meat of the correspondance.

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michaelpaulheart | eNoter

Posted February 21, 2012 at 12:09 AM (Answer #4)

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The key condition determinate of the answer above is "equally" likely: 1:16 probability. 

An unequal probability--yielding exclusively any one of the four desired results once--would obtain from a smaller number of articles in the bag (6):

  • 1 news article
  • 1 novel
  • 4 periodicals

1:4 probability, the number of possible combinations listed in the question equal to the number of possible draws from the bag.

Note that the original question did not include the conditional "exclusively"--"What is the smallest number of articles in the bag satisfying these conditions exclusively?"--admitting of additional possibilities.

Therefore--Included in the answer is the additional possibility of drawing:

  1. 4 news papers
  2. 4 novels
  3. 1 news paper and 3 novels
  4. 1 news paper and 3 periodicals
  5. 2 news papers and 2 novels
  6. 3 news papers and 1 novel
  7. 3 news papers and 1 periodical
  8. 3 novels and 1 periodical
  9. 2 news papers and 2 periodicals
  10. 2 novels and 2 periodicals
  11. 1 news paper, 2 novels and 1 periodical
  12. 2 news papers, 1 novel and 1 periodical

Add to these 12 (unexcluded possibilities) the 4 possibilities proposed in the question and the probability of drawing any one of the 4 originally proposed combinations is demonstrated to be 1:16.

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michaelpaulheart | eNoter

Posted February 21, 2012 at 6:13 AM (Answer #5)

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My apology:  I did not advert to the "hardcover" article in the original proposed question.

The absolute minimum number of articles necessary for an unequal probability of any one of the 4 possible results proposed in the original question requires (7):

  • 1 newspaper
  • 1 novel
  • 1 hardback
  • 4 periodicals

In the original question the probability of results (2) and (3) is > probability of results (1) and (4), with the probability of result (1) > (4) under the conditions listed immediately above.

Increasing order of probability then in that case seems to me to be:

  1. result (4) selection of one of each (includes the hardback)
  2. result (1) selection of 4 periodicals
  3. result (3) selection of 1 newspaper, 1 novel, and 2 periodicals
  4. result (2) selection of 1 novel and 3 periodicals.

The probability of a selection having a larger number of periodicals than 1 is greater than the probability of 1 of each of the 4 articles or the probability of 4 periodicals only, due to the greater number of the periodicals 4:1 each or 4:3 collectively in the bag.

But this unequal probability is removed with the inclusion of the same number of newspapers, of novels, of hardbacks as the number of periodicals.  Hence, the minimum number of articles providing the equal probability of each of the 4 possible selections listed in the original question is 16, 4 of each.

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