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Quiz of the Month (January 2007)

Hector C. Parr

***

SOLUTION TO LAST MONTH'S QUIZ

  1.  15
  2.  511
  3.  9330

Notes.
Q.2  Temporarily label the groups X and Y.
   Each person goes into group X or Y.
  So total number of arrangements = 210 = 1024
   But this includes cases where X or Y is empty.
  So number of arrangements = 1024-2 = 1022        -   -   A
   But on removing group labels, groups may be interchanged.
  So number of different arrangements = 1022/2 = 511

Q.3  Temporarily label the groups X, Y and Z.
   Each person goes into group X or Y or Z.
  So total number of arrangements = 310 = 59,049
   But this includes cases where one group is empty, and all
     people are distributed among other two groups
     i.e. 3*1022 = 3066 cases (see A above)
   It also includes 3 cases where two groups are empty.
  So total number of arrangements = 59,049-3066-3 = = 55,980
   On removing group labels groups may be rearranged in 6 ways.
  So number of different arrangements = 55,980/6 = 9330

THIS MONTH'S QUIZ

  A street consists of 20 houses, and the numbers of children
    living in each house are as follows:
      
      No. of children in house:         0   1   2   3   4
      No. of houses:                    4   6   5   3   2

  1. If a house is chosen at random, what is the probability
       that exactly 3 children live there?

  2. If a child is chosen at random, what is the probability that
       he or she lives in a house where exactly 3 children live?

  3. If two different houses are chosen at random, what is the
       probability that they are homes to the same number of children?
***

© Hector C. Parr (2007)

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