document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=465798>Question 465798</A>: <I><SPAN STYLE=\"word-break: break-all;\"> My homework is due tonight and I can't figure out these two problems:\r<BR>\n" );
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document.write( "5. A fair coin is flipped 20 times and the sequence of heads and tails is observed. Find the probability that the sequence consists of AT LEAST two heads. Answer in fraction reduced to the lowest terms.<BR>\n" );
document.write( "The previous answer provided was wrong:<BR>\n" );
document.write( "P(2<= x <=20) = 1 - P(0<= x <=1) = 1 - [1/2^20 + 20C2(1/2^20)]<BR>\n" );
document.write( "= 1 - [(1+190)/2^20]<BR>\n" );
document.write( "= 1 - (191/2^20)<BR>\n" );
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document.write( "== 1048385/1048576\r<BR>\n" );
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document.write( "8. Refer to the given graph. Assume one can move only down or to the right. If a route starts at A and ends at C, what is the probability that the route passes through B? Answer in fraction reduced to the lowest terms.\r<BR>\n" );
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document.write( "a<BR>\n" );
document.write( "[][][]b[][]<BR>\n" );
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document.write( "[][][][][]c\r<BR>\n" );
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document.write( "i said c(4,1) x c(6,2) = 60<BR>\n" );
document.write( "But it was wrong </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #319194 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=319194\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "You are approaching the problem correctly, but I'm not sure you have the formula appropriately used.\r<BR>\n" );
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document.write( "The probability of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large k\"> successes in <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large n\"> trials where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large p\"> is the probability of success on any given trial is given by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P_n(k,p)\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k}\">\r<BR>\n" );
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document.write( "Where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \left(n\cr k\right\)\"> is the number of combinations of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large n\"> things taken <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large k\"> at a time and is calculated by <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{n!}{k!(n\,-\,k)!}\">\r<BR>\n" );
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document.write( "but you want <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large P_{20}(\geq2,0.5)\"> which would require calculating:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P_{20}(\geq2,0.5)\ =\ \sum_{i=2}^{20}\ \left(20\cr\ i\right\)\left(0.5\right)^i\left(1\,-\,0.5\right)^{20\,-\,i}\">\r<BR>\n" );
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document.write( "A daunting task indeed. However <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 1\ -\ P_{20}(\leq1,0.5)\"> which can be rendered:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ P_{20}(\leq1,0.5)\ =\ 1\ -\ \left\[\left(20\cr\ 0\right\)\left(0.5\right)^0\left(0.5\right)^{20}\ +\ \left(20\cr\ 1\right\)\left(0.5\right)^1\left(0.5\right)^{19}\right\]\ =\ 1\ -\ 21\,\cdot\,\left(0.5\right)^{20}\ >\ 0.999\">\r<BR>\n" );
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document.write( "Is much easier to calculate.  In fact, looking a little deeper, you managed to come up with the correct denominator, but your numerator is off by a touch -- should have been 1048576 minus 21 = 1048555.\r<BR>\n" );
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document.write( "You only get one problem per posting.  But I'll give you a hint for the second one.  Since you can only move down and right, if any of your first three moves are down, you will miss point B.  Therefore your first three moves must be to the right.  From there you have a limited number of ways to go, so it should be relatively easy to count the number of successful paths.\r<BR>\n" );
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document.write( "By the way, I know just by looking at your answer of 60 that it is incorrect without looking at any other part of the problem.  Why do you think that is?  Hint:  You are trying to find a probability.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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