Dear Friends:

The questions for today are quite simple.

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?

2) How many ways can these people be arranged if they must be seated with people of the opposite sex seated next to eachother?

3) How many ways can these people be arranged if they must be positioned with people of the same sex next to each other?

Faithfully,

Douglas Castle

Answers to yesterday's quiz follow:

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A bucket contains 50 painted golf balls. 30 are painted blue, and the balance are painted red. Assume that the balls have been well shuffled around, and that you are blindfolded (in a non-hostile situation, and not for the purposes of doing anything that your sainted parents wouldn't approve of). Here are three questions:

1. What is the percentage likelihood (probability) that a ball that you choose will be red? 40%...found by subtracting 30 blue balls (ahem) from 50 in total (which leaves 20 red balls), and dividing the result by 50. 20/50= 40%

2. What is the probability that, if the first ball was red and eliminated from the game, that the next ball that you selected would be blue? In this case, we now have 49 balls in total, with 19 red and 30 blue. Dividing 30 by 49 gives us our answer, which is 61.22%.

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; if a red had been eliminated, the new probability of blue being chosen would be 30/49. Stay tuned...we'll have to save this third answer for next time. With New Year's Day fast approaching, this gives you another two days to solve this one. Invest the time wisely!

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