Page:A philosophical essay on probabilities Tr. Truscott, Emory 1902.djvu/120

110 deceives one time in ten, so that the probability of his testimony is $9⁄10$. Here the event observed is the witness attesting that number 79 is drawn. This event may result from the two following hypotheses, namely: that the witness utters the truth or that he deceives. Following the principle that has been expounded on the probability of causes drawn from events observed it is necessary first to determine à priori the probability of the event in each hypothesis. In the first, the probability that the witness will announce number 79 is the probability itself of the drawing of this number, that is to say, $1⁄1000$. It is necessary to multiply it by the probability $9⁄10$ of the veracity of the witness; one will have then $9⁄1000$ for the probability of the event observed in this hypothesis. If the witness deceives, number 79 is not drawn, and the probability of this case is $999⁄1000$. But to announce the drawing of this number the witness has to choose it among the 999 numbers not drawn; and as he is supposed to have no motive of preference for the ones rather than the others, the probability that he will choose number 79 is $1⁄999$; multiplying, then, this probability by the preceding one, we shall have $1⁄1000$ for the probability that the witness will announce number 79 in the second hypothesis. It is necessary again to multiply this probability by $1⁄10$ of the hypothesis itself, which gives $1⁄10000$ for the probability of the event relative to this hypothesis. Now if we form a fraction whose numerator is the probability relative to the first hypothesis, and whose denominator is the sum of the probabilities relative to the two hypotheses, we shall have, by the sixth principle, the probability of the first hypothesis, and