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CTK Exchange
Matt
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May-05-02, 05:17 AM (EST) |
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"Probability"
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Hey. I came across this intersting idea while reading this mathematics book but don't seem to understand it. It goes like this: A man who travels all the time was wooried about a terrorist carrying a bomb on the plane so he calculated the probabilty (whatever that it) of a terrorist coming on board with one, but still, he wasn't satisfied. So, as a solution, he now carry's a bomb on board everytime he flies because he figures two bombs on board are more rare (less probable) than one. Can someone please help me with my problem? Thank you |
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Omar
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May-05-02, 01:15 PM (EST) |
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1. "RE: Probability"
In response to message #0
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What do you see as the problem? As you might guess, considering the subject matter, that is a fairly old joke. The basis of the humor seems to be the tension between the man's apparently logical solution and the nagging feeling that he can't be right. Under the assumption that the problem to which you referred is to make that nagging feeling more concrete, I offer the following course of action. I will first, in the interest of increased portability of the puzzle in these troubled times, move the locale to a church picnic and recast the bomb as green bean and onion ring caserole. Assume that the probaility that someone will bring green bean and onion ring caserole to the church picnic is given by P(C). Assume further that you have, in self defense, brought along a green bean and onion ring caserole. What is the probability that there will be two green bean and onion ring caserole's at the picnic? You have all the information you need for the calculation. |
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Mike D
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Jun-26-02, 08:58 PM (EST) |
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2. "never leave home without it."
In response to message #0
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This is a wonderful example of conditional probability and independence. (see https://www.cut-the-knot.com/fta/Buffon/Conditionalprobability.shtml for a formal definition of conditional probability) Once we understand conditional probability, we can define independence as p(A) = p(A given B). In English, this just means that knowing event B occured doesn't give us any new information about event A. For instance, the probability of flipping a heads on a fair coin is .5. If I then tell you that Texas won the College World Series, do you want to change your estimation of the likelihood of flipping a heads on your fair coin? Most likely not, these are (hopefully) independent events. Going back to the (revised) problem, let's look at the two events, "A casserole is brought to the picnic" and "I brought a casserole to the picnic". Let's say the probability of "A casserole is brought to the picnic" is p(C). If I then tell you that I brought a casserole to the picnic, will you want to change your estimation of the likelihood that a casserole was brought to the picnic? Of course you do, you want to change it to 1, since you know for a fact that there is a casserole at the picnic. In fact, the problem boils down to semantics; the two events we should look at are actually, "Someone else brings a casserole to the picnic" and "I brought a casserole to the picnic"... If we assume these two events are independent, which is reasonably likely, then the fact that I show up with a casserole has no bearing whatsoever on the likelihood anyone else will bring a casserole. So you see, if you want to decrease the likelihood someone else brings a bomb on a plane it's just as logical to bring a green bean casserole as a bomb of your own. |
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