document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1172491>Question 1172491</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose 220 misprints are distributed randomly throughout a book of 200 pages. The probability that a given page contains, find the probability that a given page contains 2 or more misprints. (Given e^-1.1 = 0.033287) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #797545 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=797545\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a Poisson with parameter 1.1, 220/200<BR>\n" );
document.write( "Look at p(0) and p(1), and subtract that from 1 to get the answer of p(>=2)<BR>\n" );
document.write( "for p(0) it is e^(-1)*1.1^0/0!=0.33287 NOTE that value of e^(-1.1), NOT 0.033287.  It is 1/e^(1.1)<BR>\n" );
document.write( "Also, given a parameter about 1, the probability of 0 should be very high.<BR>\n" );
document.write( "p(1)=e^(-1.1)*1.1^1/1!=0.36616<BR>\n" );
document.write( "That sum is 0.699<BR>\n" );
document.write( "So the answer is 1-0.699=0.301. </SPAN>\n" );
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