Dear Friends:
Happy New Year!
There's no quiz for today. Your exercise is to invest some time in planning how to make 2009 the best year in your life's experience. I am hopeful that you will be dedicate some of this time to getting to know your inner-selves better, and to resolving to become brighter, better thinkers, and greater contributors to the Collective Body of Human Knowledge and Experience.
Faithfully,
Douglas Castle
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Answers to the last two quizzes are posted below:
Assume (although we should not assume too often) that we have two men and two women whom we wish to seat in a row:
1) How many ways can these four people be arranged? In 24 different ways. The reasoning: There are 4 people to choose from to occupy the first seat; 3 people left to occupy the second; 2 to occupy the third; and 1 to occupy the last. 4x3x2x1 = 24 different possible arrangements.
2) How many ways can these people be arranged if they must be seated with people of the opposite sex seated next to eachother? There are four configurations for seating, which are MFMF, FMFM, FMMF and MFFM (where M = male and F = female). The computations for each configuration are 2x2x1x1 (=4 arrangments), 2x2x1x1 (= 4 arrangements), 2x2x1x1 (= 4 arrangements), and 2x2x1x1 (=4 arrangements). If we add these together, we get 16 different ways.
3) How many ways can these people be arranged if they must be positioned with people of the same sex next to each other? There are four configurations for seating, which are MMFF, FFMM, FMMF and MFFM. Using the same problem-solving strategy as above, we get a total of 16 different ways.
Many people instantly solved 1) and 2), and thought that the answer to 3) could be obtained by simply subtracting 2) from 1). This is wrong because both 2) and 3) share two configurations which satisfy the rules of each arrangement. Another tricky (counter-intuitive) issue is the notion that 16 ways plus 16 ways = 32 ways. Obviously 32 is greater than 24. But when we think about it, we realize that this is indeed sensible, because of the fact that 2) includes some of the same arrangements as in 3), so there will have to be an element of double-counting.
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Our remaining problem from the previous day's quiz:
3. What is the probability that, if the first ball's color was not identified and eliminated from the game, that the next ball that you selected would blue?
Yep. A trick question. The answer would be the same as in question 1, above, because you are merely performing the same operation over again with the same number of balls. However...if you read the question thinking that a ball with an unidentified color was eliminated and not replaced, the answer is more complicated. It is what my math tutor used to call a "conditional probability problem". It poses a challenge.
If the eliminated ball were red, the number of blue remaining would be 30, with 19 red remaining; but if the eliminated ball were blue, the number of blue remaining would be 29, with 20 red left. In either case, the number of balls left over would be 49. What to do?
If a blue had been eliminated, the new probability of a blue being chosen would be 29/49 (.59); if a red had been eliminated, the new probability of blue being chosen would be 30/49 (.61). Either one of two things has occurred, and we do not know which...but we do know the likelihood (probability) of each one of these events. We know that the first one (a blue ball being chosen) was 30/50 (or .60), and that the second one (a red ball being chosen) was 20/50 (or .40).
The computation would be as follows (.60)x(.59) + (.40)x(.61) = .354 [the probability of a blue being chosen after a blue was eliminated from the group] + .244 [the probability of a blue being chosen after a red was eliminated from the group] = .598, or 59.8% . What we have done is to find the probabilities of each possibility (.60 and .40), and multiplied each by its resultant probability of blue being chosen. We added them together because we had to count both of possible ways in which we eliminated a ball from the group, i.e., by picking a blue one first, or by picking a red one first. (phew!)
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