document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1102408>Question 1102408</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 3.5 per week. Find the probability of the following events.\r<BR>\n" );
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document.write( "A. No accidents occur in one week.\r<BR>\n" );
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document.write( "B. 4 or more accidents occur in a week.\r<BR>\n" );
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document.write( "C. One accident occurs today. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #717108 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=717108\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P(0)=e-(3.5)*3.5^0/0!; and 0!=1<BR>\n" );
document.write( "This is =0.0302<BR>\n" );
document.write( "For 4 or more do 1,2,3<BR>\n" );
document.write( "P(1)=e^(-3.5)*3.5/1=0.1057.<BR>\n" );
document.write( "P(2)=e^(-3.5)*3.5^2/2!=0.1850<BR>\n" );
document.write( "P(3)=e^(-3.5)*3.5^3/3!=0.2158<BR>\n" );
document.write( "P(0,1,2,3)=0.5367<BR>\n" );
document.write( "1-0.5367 is P(4 or more)=0.4633<BR>\n" );
document.write( "Today is one day out of 7, so the Poisson parameter is 1/7 of 3.5 or 0.5<BR>\n" );
document.write( "P(1)=e-(0.5)*0.5^1/1!=0.3033 </SPAN>\n" );
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