document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1088948>Question 1088948</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A document is signed with an ElGamal signature. The probability that a signature is forged is (.1). Find the probability that at least one of the three Rob’s signatures is forged. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #703301 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=703301\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> we can use the binomial probability distribution<BR>\n" );
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document.write( "Probability (P) (k successes out of n trials) = nCk * p^k * (1-p)^(n-k)<BR>\n" );
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document.write( "Note that nCk = n! / (k! * (n-k)!), p = 0.1, n = 3<BR>\n" );
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document.write( "P (at least one of the 3 signatures is forged) = 1 - P(k=0 out of 3 trials)<BR>\n" );
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document.write( "P(k=0 out of 3 trials) = 3C0 * (0.1)^0 * (1-0.1)^(3-0) = 0.729<BR>\n" );
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document.write( "P (at least one of the 3 signatures is forged) = 1 - 0.729 = 0.271<BR>\n" );
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