On Tue, 29 Jul 2003 10:05:39 -0400, Bruce Weaver <[EMAIL PROTECTED]> wrote:
>James Lo wrote: >> Suppose that four cards are drawn successively from an ordinary deck >> of 52 cards, with replacement and at random. What is the probability >> of drawing at least one king? >> >> Answer from the book: 0.274 >> >> My solution and answer: >> >> Let AC = be the sample points where King is not found >> AC = 51*51*51*51 or 51^4 = 6765201 >> >> Let A = be the total number of sample points >> >> A = 52*52*52*52 or 52^4 = 7311616 >> >> Pr(drawing at least one king) = 1 - Pr(drawing no king) = 1 - >> (51^4/52^4) = 1 - 0.9253 = 0.0747 >> >> >> Anyone can clarify? If my answer is correct, then the answer shown on >> the textbook could be a typo error. Otherwise, which part is my error. > > >I get the same answer as your textbook. > >Had the question been about a King of a particular suit >(e.g., the King of Clubs), your solution would be correct. >But the question is about *any* King. How many Kings are >there in the deck? > >Cheers, >Bruce Can you please show me the solution? Thanks. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
