1 Ex. 3. Find the chance of throwing 10 with two dice. Ex. 4. Find the chance of throwing 15 with three dice. Ex. 5. A and B each throws a die; shew that it is 7: 5 that A's throw is not greater than B's. Ex. 6. A and B each throw with two dice: find the chance that their throws are equal. 73 Ans. 648 Ex. 7. A and B have equal chances of winning a single game at tennis: find the chance of A winning the 'set' (1) when A has won 5 games and B has won 4, (2) when A has won 5 games and B has won 3, and (3) when A has won 4 games and B has won 2. 3 7 13 Ans (1) (2) (3) 4' 8' 16 Ex. 8. A and B have equal chances of winning a single game; and A wants 2 games and B wants 3 games to win a match: shew that it is 11 to 5 that A wins the match. Ex. 9. A and B have equal chances of winning a single game; and A wants n games and B wants n+1 games to win a match: shew that 1.3.5... (2n-1) 1.3.5... (2n-1) 2.4.6...2n 2.4.6...2n to 1 the odds on A are 1 + 3 Ex. 10. A's chance of winning a single game against B is chance of his winning at least 2 games out of 3. : find the 5 81 Ans. 125* 2 Ex. 11. A's chance of winning a single game against B is :: find the 3་ chance of his winning at least 3 games out of 5. Ex. 12. What is the chance of throwing at least 2 sixes in 6 throws with a die? Ans. 12281 46656* Ex. 13. A coin is tossed five times in succession: shew that it is an even chance that three consecutive throws will be the same. Ex. 14. Three men toss in succession for a prize which is to be given to the first who gets 'heads'. Find their respective chances. 405. The value of a given chance of obtaining a given sum of money is called the expectation. is the chance of obtaining a sum of money M; For if E be the expectation in one trial, E (a+b) will be the expectation in a + b trials. But the chance being a the sum M will, on the average, be won a times in a+b' every a+b trials; and hence the expectation in a+b trials is Ma. Hence E (a + b) = Ma; therefore Thus the expectation is the sum which may be won multiplied by the chance of winning it. Ex. 1. A bag contains 5 white balls and 7 black ones. Find the expectation of a man who is allowed to draw a ball from the bag and who is to receive one shilling if he draws a black ball, and a crown if he draws a white one. The chance of drawing a black ball is 5 7 12; and therefore the The chance of drawing expectation from drawing a black ball is 7d. Ex. 2. A purse contains 2 sovereigns, 3 half-crowns and 7 shillings. Ex. 3. Two persons toss a shilling alternately on condition that the first who gets 'heads' wins the shilling: find their expectations. Ans. 8d., 4d. Ex. 4. Two persons throw a die alternately, and the first who throws 6 is to receive 11 shillings: find their expectations. Ans. 6s., 5s. 406. Inverse Probability. When it is known that an event has happened and that it must have followed from some one of a certain number of causes, the determination of the probabilities of the different possible causes is said to be a problem of inverse probability. For example, it may be known that a black ball was drawn from one or other of two bags, one of which was known to contain 2 black and 7 white balls and the other 5 black and 4 white balls; and it may be required to determine the probability that the ball was drawn from the first bag. Now, if we suppose a great number, 2N, of drawings to be made, there will in the long run be N from each bag. But in N drawings 2 from the first bag there are, on the average, 9 N which give a black 5 and in N drawings from the second bag there are N which 9 9 ball; give a black ball. Hence, in the long run, N out of a total of 2 5 N+ 9 9 N black balls are due to drawings from the first bag; thus the probability that the ball was drawn from the first bag is Let P., P. P1 be the probabilities of the existence of n causes, which are mutually exclusive and are such that a certain event must have followed from one of them; and let P1 P2 be the respective probabilities that when one of the causes P1, P2 P exists it will be followed by the event in question; then on any occasion when the event is known to have occurred the probability of the rth cause is ין .... Pn n 1 Let a great number N of trials be made; then the first cause will exist in N. P1 cases, and the event will follow in N. P1· P1 cases. So also the second cause exists and the event follows in N. P2.P2 cases; and so on. 2 Hence the event is due to the rth cause in N.Pr. Pr S. A. 33 cases out of a total of N (P,p,+ P2P2+ ... + P„Pn); the probability of the rth cause is therefore P.P ΣPp' Having found the probability of the existence of each of the different causes, the probability that the event would occur on a second trial can be at once found. is For let P be the probability of the existence of the rth cause; then p, is the probability that the event will happen when the rth cause exists; and therefore P. the probability that the event will happen from the rth cause. .Pr Hence, as the causes are mutually exclusive, the probability that the event would happen on a second trial is P1 · P1 + P2 · P2 + ... + Pn' · Pn' 2 Ex. 1. There are 3 bags which are known to contain 2 white and 3 black, 4 white and 1 black, and 3 white and 7 black balls respectively. A ball was drawn at random from one of the bags and found to be a black ball. Find the chance that it was drawn from the bag containing the most black balls. Ex. 2. From a bag which is known to contain 4 balls each of which is just as likely to be black as white, a ball is drawn at random and found to be white. Find the chance that the bag contained 3 white and 1 black balls. 16' The bag may have contained (1) 4 white, (2) 3 white and 1 black, (3) 2 white and 2 black, (4) 1 white and 3 black, and (5) 4 black; and the chances of these are respectively 16' 16' 16' 16 1 4 6 4 1 and Art. 404. Also the chances of drawing a white ball in these 3 1 1 different cases will be 1, 4' 2' 4 and O respectively. 407. Probability of testimony. The method of dealing with questions relating to the credibility of witnesses will be seen from the following examples: Ex. 1. A ball has been drawn at random from a bag containing 99 black balls and 1 white ball; and a man whose statements are accurate 9 times out of 10 asserts that the white ball was drawn. Find the chance that the white ball was really drawn. The probability that the white ball will really be drawn in any case 1 is and therefore the probability that the man will truly assert 100 that the white ball is drawn is 1 9 99 The probability that the white ball will not be drawn is and 100' therefore the probability that the man will falsely assert that the white ball is drawn is X 99 1 100 10 Hence as in Art. 406 the required probability is 100 Ex. 2. From a bag containing 100 tickets numbered 1, 2, respectively, a ticket has been drawn at random; and a witness, whose statements are accurate 9 times out of 10, asserts that a particular ticket has been drawn. Find the chance that this ticket was really drawn. In 1000 N trials the ticket in question will be drawn 10 N times; and the witness will correctly assert that it has been drawn 9 N times. The ticket will not be drawn in 990 N cases, and the witness will make a wrong assertion in 99 N of these cases; but there are 99 ways of making a wrong assertion and these may all be supposed to be equally likely; hence the witness will wrongly assert that the particular ticket has been drawn in N cases. Hence the required 9 " 10 probability is so that the probability is in this case equal to the probability of the witness speaking the truth. Ex. 3. A speaks the truth three times out of four, and B five times out of six; and they agree in stating that a white ball has been drawn from a bag which was known to contain 1 white and 9 black balls. Find the chance that the white ball was really drawn. The probability that the white ball will be drawn in any case is |