document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=535145>Question 535145</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose 200 misprints ar distributed randomly throughout a book of 600 pages.  Find the probability that a given page contains.<BR>\n" );
document.write( "(a) exactly two misprints<BR>\n" );
document.write( "(b) two or more misprints </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #351843 by </B><A HREF=/tutors/aboutme.mpl?userid=SwiftAlbatross><B>SwiftAlbatross(13)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=SwiftAlbatross><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=SwiftAlbatross><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=351843\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assuming the chance of a misprint being on any page is uniform and that each \r<BR>\n" );
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document.write( "misprint is an independent event, we can think of a misprint being randomly \r<BR>\n" );
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document.write( "assorted into 600 pages and this occurrence happening 200 times. Therefore, the \r<BR>\n" );
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document.write( "probability a misprint will go to any page is 1/600 since there are 600 pages and \r<BR>\n" );
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document.write( "a misprint may go into any of them evenly. Then, like flipping coins, the next \r<BR>\n" );
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document.write( "misprint event happens independently of the previous one and so its probability of \r<BR>\n" );
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document.write( "going to any page is also 1/600, and so on and so on for all 200 misprints. Then, \r<BR>\n" );
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document.write( "just like coins, you can multiply their probabilites together. For example, what \r<BR>\n" );
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document.write( "is the probability of flipping 3 Heads in 3 coin tosses: Simply (1/2)*(1/2)*(1/2). \r<BR>\n" );
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document.write( "The same logic can be used here: For instance, the probability that one page will \r<BR>\n" );
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document.write( "have all 200 misprints is (1/600)^200.<BR>\n" );
document.write( " Part (a) asks that there are exactly 2 misprints on one page and the other 198 \r<BR>\n" );
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document.write( "misprints must be on the other 599 pages. So you can do <BR>\n" );
document.write( "(1/600) * (1/600) * (599/600)^198 * 200C2<BR>\n" );
document.write( "where 200C2 is 200 Choose 2 or C(200,2) [Combinations: I'll explain later]\r<BR>\n" );
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document.write( "Part (b) asks that one page contains 2 or more misprints. We could calculate the \r<BR>\n" );
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document.write( "probability of a page getting exactly 2 misprints and then add to it the \r<BR>\n" );
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document.write( "probability of getting exactly 3 misprints and then add the probability of getting \r<BR>\n" );
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document.write( "exactly 4 misprints and so on, but that is time consuming so we can just subtract \r<BR>\n" );
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document.write( "the probability of getting exactly one misprint on that page and exactly no misprints on  \r<BR>\n" );
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document.write( "that page from 1, since this is the same probability. So\r<BR>\n" );
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document.write( "1 - (1/600) * (599/600)^199 * 200C1 - (599/600)^200 * 200C0\r<BR>\n" );
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document.write( "To convince yourself of this logic, use smaller numbers and work with actual \r<BR>\n" );
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document.write( "paper. For example, suppose you have only 6 pages and 2 misprints going among \r<BR>\n" );
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document.write( "them. A misprint may go to the 6 pages evenly. So the chance page 1 will have 2 \r<BR>\n" );
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document.write( "misprints is :\r<BR>\n" );
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document.write( "(1/6) * (1/6)<BR>\n" );
document.write( "The first misprint must go to page 1 AND the second misprint must go to page 1.\r<BR>\n" );
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document.write( "exactly one misprint:\r<BR>\n" );
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document.write( "(1/6) * (5/6) + (5/6) * (1/6)<BR>\n" );
document.write( "The first misprint must go to page 1 AND the second misprint must go to pages 2-6<BR>\n" );
document.write( "OR<BR>\n" );
document.write( "The first misprint must go to pages 2-6 AND the second misprint must go to page 1\r<BR>\n" );
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document.write( "no misprints:\r<BR>\n" );
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document.write( "(5/6) * (5/6)<BR>\n" );
document.write( "The first misprint must go to pages 2-6 AND the second misprint must go to pages 2-6.\r<BR>\n" );
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document.write( "one or more misprints:<BR>\n" );
document.write( "(1/6) * (5/6) + (5/6) * (1/6) + (1/6) * (1/6) = 1- (5/6) * (5/6)<BR>\n" );
document.write( "Either write the case for exactly one misprint plus the case for exactly two \r<BR>\n" );
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document.write( "misprints or subtract the case for exactly no misprints from one. \r<BR>\n" );
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document.write( "Even though I write first and second misprint, they are independent events so the order of \r<BR>\n" );
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document.write( "the misprints is inconsequential but to keep track of which misprint goes where, I \r<BR>\n" );
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document.write( "order them.\r<BR>\n" );
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document.write( "Now imagine you have four misprints in 6 pages.\r<BR>\n" );
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document.write( "P(Exactly 4 misprints on a given page) =<BR>\n" );
document.write( "(1/6) * (1/6) * (1/6) * (1/6)\r<BR>\n" );
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document.write( "P(Exactly 3 misprints on a give page) =<BR>\n" );
document.write( "(1/6)*(1/6)*(1/6)*(5/6) + (1/6)*(1/6)*(5/6)*(1/6) + (1/6)*(5/6)*(1/6)*(1/6) +<BR>\n" );
document.write( "(5/6)*(1/6)*(1/6)*(1/6) = (1/6)*(1/6)*(1/6)*(5/6)*4\r<BR>\n" );
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document.write( "P(Exactly 2 misprints on a given page) =<BR>\n" );
document.write( "(1/6)*(1/6)*(5/6)*(5/6) + (1/6)*(5/6)*(1/6)*(5/6) + (1/6)*(5/6)*(5/6)*(1/6) + <BR>\n" );
document.write( "(5/6)*(1/6)*(1/6)*(5/6) + (5/6)*(1/6)*(5/6)*(1/6) + (5/6)*(5/6)*(1/6)*(1/6) =<BR>\n" );
document.write( "6 * (1/6)*(1/6)*(5/6)*(5/6)\r<BR>\n" );
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document.write( "As you can see, the number being multiplied is just the differnt combinations of \r<BR>\n" );
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document.write( "misprints appearing from the total number of misprints. So nCr = (# of total \r<BR>\n" );
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document.write( "misprints) C (# of misprints appearing on target page). For the case of exactly 2 \r<BR>\n" );
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document.write( "misprints on a given page that preceded, this is 4C2 or 6. </SPAN>\n" );
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