Comments embedded in original post, below: On Tue, 29 Jul 2003, James Lo wrote (edited):
> In tossing four fair dice, what is the probability of tossing at > most one 3? [Answer : 0.868] > My solution and answer: > Let A = total number of sample points << O.K. > B = sample points with at most one 3 << O.K. > Pr(B/A) = 5*6*6*6 / 6*6*6*6 = 1080 / 1296 = 0.8333 > Where is my error? << In logic. B is not = 1080. You have calculated the number of points for which one die does NOT show a 3; so you have imposed that restriction, and at the same time permitted the other 3 dice to take on any value INCLUDING 3. 1080 thus includes all sample points for which the number of 3's is zero, and SOME of the sample points where the number of 3's is 1 or 2 or 3. Think about it again. ----------------------------------------------------------------------- Donald F. Burrill [EMAIL PROTECTED] 56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816 . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
