Publisher's Synopsis
This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1867 edition. Excerpt: ...among balls of the same colour when we calculated that number. (See Choice, page 33.). Question.--In a bag are five red balls, seven white balls, four green balls, and three black balls. If they be drawn one by one, what is the chance that all the red balls should be drawn first, then all the white ones, then all the green ones, and then all the black ones? Answer.--The nineteen balls can be arranged in 119 5. 7. 4. 3 different orders (Choice, Rule VII.). All these are equally likely, and therefore the chance of any particular order is 5.7.4.8 This will be the chance required, for all individuality among balls of the same colour has been disregarded; only one of the different arrangements will give the order of colours prescribed in the question. Question.--Out of a bag containing 12 balls, 5 are drawn and replaced, and afterwards 6 are drawn. Find the chance that exactly 3 balls were common to the two drawings. Anstcer.--The second drawing could be made altogether in 12----, or 924 6 6 ways. Bnt it could be made so as to include exactly 3 of the balls contained in the first drawing in 32 X3Tr35 ways; for it must consist of a selection of 3 balls out of the first 5, and a selection of 3 balls out of the remaining 7 (Choice, Rules Viii. and EC.). Hence, the chance that the second drawing should contain exactly 3 balls common to the first, is-rr-or. As the respective probabilities of various throws, with two common dice, are of practical interest, in their bearing upon such games as Backgammon, it may be well to discuss this case with some completeness. It will be observed that as each die can fall in six ways, the whole number of ways in which the two dice can fall is 6 x 6 or 36. But these 36 different ways are not practically different...