CHAPTER XXXII. PROBABILITY. 449. DEFINITION. If an event can happen in a ways and fail in b ways, and each of these ways is equally likely, the probability, or the chance, of its happening is α a + b' and that of its failing is b a+b' For instance, if in a lottery there are 7 prizes and 25 blanks, 7 the chance that a person holding 1 ticket will win a prize is 32' 450. The reason for the mathematical definition of probability may be made clear by the following considerations: If an event can happen in a ways and fail to happen in b ways, and all these ways are equally likely, we can assert that the chance of its happening is to the chance of its failing as a to b. Thus if the chance of its happening is represented by ka, where k is an undetermined constant, then the chance of its failing will be represented by kb. .. chance of happening + chance of failing = k (a + b). Now the event is certain to happen or to fail; therefore the sum of the chances of happening and failing must represent certainty. If therefore we agree to take certainty as our unit, we have 1 1=k(a+b), or k= ... the chance that the event will happen is α a+ and the chance that the event will not happen is b a+b' COR. If p is the probability of the happening of an event, the probability of its not happening is 1 – p. 451. Instead of saying that the chance of the happening of an event is α it is sometimes stated that the odds are a to b a + b' in favour of the event, or b to a against the event. 452. The definition of probability in Art. 449 may be given in a slightly different form which is sometimes useful. If c is the total number of cases, each being equally likely to occur, and of these a are favourable to the event, then the probability that the event will happen is, and the probability that it will not C Example 1. What is the chance of throwing a number greater than 4 with an ordinary die whose faces are numbered from 1 to 6? There are 6 possible ways in which the die can fall, and of these two are favourable to the event required; Example 2. From a bag containing 4 white and 5 black balls a man draws 3 at random; what are the odds against these being all black? and The total number of ways in which 3 balls can be drawn is 9C3, the number of ways of drawing 3 black balls is 5C3; therefore the chance of drawing 3 black balls = 5C, 5.4.3 5 3 = 9C39.8.7 42 Thus the odds against the event are 37 to 5. Example 3. Find the chance of throwing at least one ace in a single throw with two dice. The possible number of cases is 6 × 6, or 36. An ace on one die may be associated with any of the 6 numbers on the other die, and the remaining 5 numbers on the first die may each be associated with the ace on the second die; thus the number of favourable cases is 11. There are 5 ways in which each die can be thrown so as not to give an ce; hence 25 throws of the two dice will exclude aces. That is, the chance not throwing one or more aces is 25 36 ; so that the chance of throwing one at least is 1 or Example 4. Find the chance of throwing more than 15 in one throw with 3 dice. A throw amounting to 18 must be made up of 6, 6, 6, and this can occur in 1 way; 17 can be made up of 6, 6, 5 which can occur in 3 ways; 16 may be made up of 6, 6, 4 and 6, 5, 5, each of which arrangements can occur in 3 ways. Therefore the number of favourable cases is 1+3+3+3, or 10. And the total number of cases is 63, or 216; 10 5 therefore the required chance= = Example 5. A has 3 shares in a lottery in which there are 3 prizes and 6 blanks; B has 1 share in a lottery in which there is 1 prize and 2 blanks: shew that A's chance of success is to B's as 16 to 7. the sum of these numbers is 64, which is the number of ways in which A can win a prize. Also he can draw 3 tickets in 9.8.7 " 1.2.3 or 84 ways; Or we might have reasoned thus: A will get all blanks in 453. Suppose that there are a number of events A, B, C,..., of which one must, and only one can, occur; also suppose that a, b, c,... are the numbers of ways respectively in which these events can happen, and that each of these ways is equally likely to occur; it is required to find the chance of each event. The total number of equally possible ways is a+b+c+ and of these the number favourable to A is a; hence the chance 454. From the examples we have given it will be seen that the solution of the easier kinds of questions in Probability requires nothing more than a knowledge of the definition of Probability, and the application of the laws of Permutations and Combinations. EXAMPLES. XXXII. a. 1. In a single throw with two dice find the chances of throwing (1) five, (2) six. 2. From a pack of 52 cards two are drawn at random; find the chance that one is a knave and the other a queen. 3. A bag contains 5 white, 7 black, and 4 red balls: find the chance that three balls drawn at random are all white. 4. If four coins are tossed, find the chance that there should be two heads and two tails. 5. One of two events must happen: given that the chance of the one is two-thirds that of the other, find the odds in favour of the other. 6. If from a pack four cards are drawn, find the chance that they will be the four honours of the same suit. 7. Thirteen persons take their places at a round table, shew that it is five to one against two particular persons sitting together. 8. There are three events A, B, C, one of which must, and only one can, happen; the odds are 8 to 3 against A, 5 to 2 against B: find the odds against C. 9. Compare the chances of throwing 4 with one die, 8 with two dice, and 12 with three dice. 10. In shuffling a pack of cards, four are accidentally dropped; find the chance that the missing cards should be one from each suit. 11. A has 3 shares in a lottery containing 3 prizes and 9 blanks; has 2 shares in a lottery containing 2 prizes and 6 blanks: compare r chances of success. 2. Shew that the chances of throwing six with 4, 3, or 2 dice ectively are as 1 : 6:18. |