document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=596214>Question 596214</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A die is rolled 15 times. What is the probability of at least two 5's appearing? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #377600 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=377600\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the binomial probability distribution to compute the probability of getting zero fives (ie P(X = 0)) and the probability of getting one five (ie P(X = 1)) .\r<BR>\n" );
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document.write( "P(X = k) = (n ncr k)*(p)^(k)*(1-p)^(n-k) <BR>\n" );
document.write( "P(X = 0) = (15 ncr 0)*(0.1667)^(0)*(1-0.1667)^(15-0) <BR>\n" );
document.write( "P(X = 0) = (15 ncr 0)*(0.1667)^(0)*(0.8333)^(15-0) <BR>\n" );
document.write( "P(X = 0) = (1)*(0.1667)^(0)*(0.8333)^15 <BR>\n" );
document.write( "P(X = 0) = (1)*(1)*(0.064866) <BR>\n" );
document.write( "P(X = 0) = 0.064866\r<BR>\n" );
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document.write( "P(X = k) = (n ncr k)*(p)^(k)*(1-p)^(n-k) <BR>\n" );
document.write( "P(X = 1) = (15 ncr 1)*(0.1667)^(1)*(1-0.1667)^(15-1) <BR>\n" );
document.write( "P(X = 1) = (15 ncr 1)*(0.1667)^(1)*(0.8333)^(15-1) <BR>\n" );
document.write( "P(X = 1) = (15)*(0.1667)^(1)*(0.8333)^14 <BR>\n" );
document.write( "P(X = 1) = (15)*(0.1667)*(0.077843) <BR>\n" );
document.write( "P(X = 1) = 0.194646\r<BR>\n" );
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document.write( "In short, \r<BR>\n" );
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document.write( "P(X = 0)= 0.064866\r<BR>\n" );
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document.write( "P(X = 1)= 0.194646\r<BR>\n" );
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document.write( "Now add the two probabilities \r<BR>\n" );
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document.write( "P(X = 0)+P(X = 1) = 0.064866 + 0.194646\r<BR>\n" );
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document.write( "P(X = 0)+P(X = 1) = 0.259512\r<BR>\n" );
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document.write( "So P(X <= 1) = 0.259512\r<BR>\n" );
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document.write( "Since the binomial distribution is a discrete distribution, this means that \r<BR>\n" );
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document.write( "P(X >= 2) = 1 - P(X <= 1)\r<BR>\n" );
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document.write( "So this means...\r<BR>\n" );
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document.write( "P(X >= 2) = 1 - P(X <= 1)\r<BR>\n" );
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document.write( "P(X >= 2) = 1 - 0.259512\r<BR>\n" );
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document.write( "P(X >= 2) = 0.740488\r<BR>\n" );
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document.write( "So the probability of at least two 5's appearing is approximately <font color=\"red\">0.740488</font>\r<BR>\n" );
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