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NeurobicsEven OdderBy Russell Dear From the correspondence received about the "car and goats" problem [July 1992], I think we are agreed that probability issues can be counter-intuitive. It was really a question of confusing probability and conditional probability. The probability, for example, that your friend has two daughters may be a quarter, but if you already know that one child was female then the probability of two daughters is one-third. The extra information changes the conditions of the problem. In the car and goats problem, the game host knew which door hid the car and took that into account when s/he chose which door to open. More information was available and the conditions of the problem changed. Those who still feel uncomfortable can settle any doubts by playing the analogous "pea under cup" game. Take one pea and three egg cups. The pea corresponds to the car in the original problem. While your back is turned, ask a friend to place the pea under one of the cups. Guess where the pea is hidden by choosing a cup (don't overturn the cup yet). Your friend, who knows where the pea is, should then nominate a cup which does not hide the pea. Change your choice and turn over the egg cup. Play the game enough times for the proportions to "firm up" -- 30 should be enough. You will find that you have chosen correctly about two-thirds of the time. While we are still on the subject of probability, here is another paradox which you may find interesting: Two drugs were tested for treatment of a certain disease. Drug A was used on a total of 700 people; 600 smokers and 100 non-smokers. It was successful with 300 non-smokers and 8 smokers. Drug B was tested on 1000 people; 200 non-smokers and 800 smokers. It was successful with 196 non-smokers and 104 smokers. Which drug is most successful in treating the disease? Compiling a table might be helpful. Which drug is the most successful? When the two groups, smokers and non-smokers are considered separately, drug B is the most efficacious (in both groups). When the data is pooled, the opposite seems to be the case! Don't think such data is necessarily contrived -- situations like this occur regularly. By the way, one of the non-smokers above had only just given up the habit. At a party recently, she was offered a cigarette. She took it and, to make a point, broke it into two pieces. The problem is, how long would you expect the smaller piece to be? (In probability, the word "expect" is specifically defined. Here it may be loosely regarded as meaning "on the average". Since on average a single six-sided die gives a score of 3.5, its expected score is 3.5.) SolutionYou may not be convinced that enough information has been given to solve this little problem. Assuming, not unrealistically, that the break in the cigarette could be anywhere, then the length of the smaller piece will lie between zero and half the length of the original cigarette. The expected length of the smaller piece is a quarter of the cigarette length. Russell Dear is a Mathematician living in Invercargill |
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