document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1195322>Question 1195322</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 19 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution.\r<BR>\n" );
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document.write( "Find the probability that at least one particle arrives in a particular one second period. Round your answer to four decimals.<BR>\n" );
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document.write( "Find the probability that at least two particles arrive in a particular 2 second period. Round your answer to four decimals. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #827919 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=827919\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> since the number is proportional to time, it would be 19/60 per second, and that is the specific Poisson parameter.<BR>\n" );
document.write( "For a given second, it is e^(-19/60)*(19/60)^1/1!=0.2307<BR>\n" );
document.write( "For two seconds, it would be parameter 38/60 or 19/30<BR>\n" );
document.write( "this is e^(-19/30)*(19/30)^2/2!=0.1065, less likely because need two relatively uncommon events in a second rather than 1, but there is a little more time, just not enough to deal with the fact the number has doubled. </SPAN>\n" );
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