document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1022437>Question 1022437</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A coin is flipped eight times in succession. In how many ways can at least five heads occur?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #638134 by </B><A HREF=/tutors/aboutme.mpl?userid=mathmate><B>mathmate(429)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mathmate><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=mathmate><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=638134\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  <BR>\n" );
document.write( "Question:<BR>\n" );
document.write( "A coin is flipped eight times in succession. In how many ways can at least five heads occur?<BR>\n" );
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document.write( "Solution:<BR>\n" );
document.write( "Assume that a fair coin is used, so P(H)=P(T)=0.5.<BR>\n" );
document.write( "We can then apply the binomial coefficients for r=5,6,7,8 in<BR>\n" );
document.write( "P(r)=C(n,r)/2^n<BR>\n" );
document.write( "where C(n,r) is the binomial coefficient (combinations) of r objects out of n.<BR>\n" );
document.write( "The above equation is a special case of the binomial distribution where the probabilities of success and failure are both 0.5.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "For r=5 to 8 heads, we need to sum the coefficients:<BR>\n" );
document.write( "P(5)+P(6)+P(7)+P(8)<BR>\n" );
document.write( "=0.21875+0.10938+0.03125+0.00391<BR>\n" );
document.write( "=0.3633<BR>\n" );
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document.write( "It is interesting to note that in this particular case where P(H)=P(T), the distribution is symmetrical, so that<BR>\n" );
document.write( "P(0)=P(8),<BR>\n" );
document.write( "P(1)=P(7),<BR>\n" );
document.write( "P(2)=P(6),<BR>\n" );
document.write( "P(3)=P(5).<BR>\n" );
document.write( "Therefore, P(5)+P(6)+P(7)+P(8)=P(0)+P(1)+P(2)+P(3)<BR>\n" );
document.write( "Since P(0)+P(1)+P(2)....+P(8)=1.0, we have<BR>\n" );
document.write( "P(5)+P(6)+P(7)+P(8)<BR>\n" );
document.write( "=(1-P(4))/2  ..... which is much simpler to calculate than summing<BR>\n" );
document.write( "=(1-0.27344)/2<BR>\n" );
document.write( "=0.3633 as before. </SPAN>\n" );
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