Page:A philosophical essay on probabilities Tr. Truscott, Emory 1902.djvu/126

116 occur in the present case when the witness does not deceive and is not mistaken at all, and when he deceives and is mistaken at the same time. One may form the four following hypotheses:

1st. The first and second witness speak the truth. Then a white ball has at first been drawn from the urn A, and the probability of this event is $1⁄2$, since the ball drawn at the first draw may have been drawn either from the one or the other urn. Consequently the ball drawn, placed in the urn B, has reappeared at the second draw; the probability of this event is $1⁄1000001$, the probability of the fact announced is then $1⁄2000002$. Multiplying it by the product of the probabilities $9⁄10$ and $9⁄10$ that the witnesses speak the truth one will have $81⁄200000200$ for the probability of the event observed in this first hypothesis. 2d. The first witness speaks the truth and the second does not, whether he deceives and is not mistaken or he does not deceive and is mistaken. Then a white ball has been drawn from the urn A at the first draw, and the probability of this event is $1⁄2$. Then this ball having been placed in the urn B a black ball has been drawn from it: the probability of such drawing is $1000000⁄1000001$; one has then $1000000⁄2000002$ for the probability of the compound event. Multiplying it by the product of the two probabilities $9⁄10$ and $1⁄10$ that the first witness speaks the truth and that the second does not, one will have $9000000⁄200000200$ the probability for the event observed in the second hypothesis.

3d. The first witness does not speak the truth and the second announces it. Then a black ball has been drawn from the urn B at the first drawing, and after