<text><span class="style10">robability: Chance and Choice (2 of 2)</span><span class="style7"></span><span class="style10">Rules of probability</span><span class="style7">If, we wish to find the combined probability of two independent trials, we use the </span><span class="style26">multiplication rule</span><span class="style7">. When we throw a pair of dice, we can consider these as two independent trials - because how one die falls does not affect the other. No matter what the first die shows, the second can show any of its six faces - and there are six ways the first can fall. So there are 6 x 6 = 36 possible outcomes for the ordered pair. Since these are equiprobable, the probability of any particular outcome, say a 1 with the first die and a 4 with the second, is 1/36, which is 1/6 x 1/6; that is, we multiply the individual probabilities to get the probability of a particular ordered outcome using the two dice. Since there are six ways of throwing a double, the probability of throwing the same number with two dice is 6/36 = 1/6, whence the probability of throwing different numbers is 1 - 1/6 = 5/6.When a third die is thrown, there are now only four available faces differing from those of the first two dice, so that the probability of this showing a third different number is 4/6. Hence, the probability of throwing three different numbers with three dice is 5/6 x 4/6. Throwing six dice, the probability of an outcome of six different numbers is 5/6 x 4/6 x 3/6 x 2/6 x 1/6, which equals 5/324, or approximately 0.015. Thus we can expect the outcome to occur only once or twice in every hundred trials.However, if we wish to specify the order beforehand (say, 1, 2, 3, 4, 5, 6 or 6, 4, 2, 5, 3, 1, for example), the probability is 1/6 for the first throw multiplied by 1/6 for the second throw, and so on. With the six dice the probability is therefore (1/6) to the power of 6 which equals 1/46656, or approximately 0.000021. Thus, we can expect such an outcome only about twice in every 100 000 trials.It is very important that we correctly define a problem before applying the addition or multiplication rule. Many problems in fact require both. Suppose, for example, we wish to achieve a total of 8 with two dice. This could be from 6 and 2, 5 and 3, or 4 and 4; but there are two other possibilities: 2 and 6, and 5 and 3. That is, there are two ways of throwing a pair of distinct numbers, so that the probability that one die will show a 6 and the other a 2 is 2/36. Similarly, for a 5 and a 3. But there is only one way of throwing a double, so that the probability of a double 4 is only 1/36. The probability of an outcome of a total of 8 is the sum of these probabilities, that is, 5/36.</span><span class="style10">Decision-making</span><span class="style7">Very often we are required to make decisions based on only a little knowledge of the likely circumstances. As examples: a doctor may have to choose between different treatments for a patient with relatively little experimental evidence as to which is more successful; or the directors of a business may have to choose between different advertising strategies on the basis of rival claims about the effectiveness of the different media. In such cases, the decision-makers need ways of measuring the competing strategies against one another. One way of doing this involves calculating the </span><span class="style26">expected value</span><span class="style7">.A simple way to explain this is by considering a hockey or soccer league table, where 2 points are awarded for a win, 1 for a draw, and 0 for a loss. Suppose a particular team in the league decides, at the beginning of the season, that - based on all the available evidence - its probability of winning any match is 1/4 and of drawing is 1/3. Then its probability of losing is 1 - 1/4 - 1/3 = 5/12. Over a run of 12 matches the team would expect to win 3, draw 4, and lose 5. The points it would expect to gain in 12 average matches would be (3 x 2) + (4 x 1) + (5 x 0) = 10 points. Thus the average number of points they can expect in each match is 10/12. This is the expected value.By calculating an expected value for each course of action available to us, we are able to choose the one that has the best likely outcome. On the basis of only partial knowledge, we are enabled to make a rational choice, even though it may not be the one with the highest probability.EJB</span><span class="style10">SHARING A BIRTHDAY</span><span class="style7">Suppose we are looking for a pair of people who have the same birthday. What is the smallest number of people, chosen at random, for which there is a better than evens chance of two sharing a birthday? Since, allowing for leap years, there are 366 possible birthdays, many people guess that 183 is the answer. In fact, the answer is 23. The probability of the second person we ask not matching the birthday of the first, is 365/366. The probability of the third person not matching either of the other two is 364/366. So the chance of three people not sharing the same birthday is 365/366 x 364/366. For n people, the chances of them all having different birthdays is, thus, 365/366 x 364/366 x 363/366 x .... to n - 1 terms. We need to find how many terms of this sequence we need to multiply together before their product becomes less than 1/2 - that is, before there is a less than evens chance that this number of people do not include a single pair with the same birthday.If we work it out, we find that the probability of 22 people having different birthdays is 0.5252, and for 23 people, it is 0.494. Therefore, 23 is the smallest number of people for which there is a better than evens chance of at least one shared birthday. On the other hand, we need 367 to be certain that two of them have the same birthday! </span><span class="style10">RATIONAL CHOICE</span><span class="style7">When my car breaks down, my garage in forms me that the cause is either the gearbox or the final drive, with odds of 3 to 2 on the gearbox being the problem. The cost of repairing the gearbox will be $200 and the final drive $150, both costs including dismantling and reassembly charges. However, if the component they examine first is undamaged, the cost of stripping it down and reassembling is $60 for the gearbox and $30 for the final drive, in addition to the cost of the repair. Where should they begin? Suppose they examine the gearbox first. In the long run, on 3 occasions out of 5 the fault will actually be there, and the cost will be $200. Otherwise, and hence with a probability of 2/5, the cost will be $60 for inspecting the gearbox and $150 for repairing the final drive - a total of $210. Thus, the expected cost of the strategy of examining the gearbox first is (3/5 x $200) + (2/5 x $210) = $204.Suppose, however, they decide to examine the final drive first. In the long run, there will be no fault on 3 occasions out of 5, incurring a cost of $30 for inspecting the drive plus $200 for repairing the gearbox. On the 2 out of 5 occasions that the fault is actually in the final drive, the cost will only be $150. Thus, the expected cost for this strategy is (3/5 x $230) + (2/5 x $150) = $198.It is thus better if the garage examines the final drive first, even though the fault is more likely to be in the gearbox. This is because this strategy has the lower expected cost. Of course, this will be of little consolation to me if the fault is in the gearbox and I have to pay $230.The garage, on the other hand, might decide to make a standard fixed charge of $230 for the job. It would still be better for the garage to examine the drive first, since - over a run of repairs - they would make an average gain of $32, whereas if they investigated the more probable cause first, their average gain would be only $26.</span></text>
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<text><span class="style10"> roulette wheel.</span><span class="style7"> On a normal roulette wheel there are 37 compartments numbered from 0 to 36. The probability of, say, 0 winning is 1/37; the odds against 0 winning are 36 to 1. The casino will not usually offer odds better than 35 to 1, so in the long run the casino will make a profit.</span></text>
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<text>ΓÇó MATHEMATICS AND ITS APPLICATIONSΓÇó NUMBER SYSTEMS AND ALGEBRA</text>
