document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1079024>Question 1079024</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a factory there are three<BR>\n" );
document.write( "independently operating machines, each with a<BR>\n" );
document.write( "probability of 0.6 of being working, find the probability that:\r<BR>\n" );
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document.write( "a. All the machines are working<BR>\n" );
document.write( "b. None of the machines are working<BR>\n" );
document.write( "c. Only one machine is working </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #693402 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=693402\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a.  Probability(P) (all three are working) = 0.6^3 = 0.216<BR>\n" );
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document.write( "b:  P(no machines are working) = 0.4^3 = 0.064<BR>\n" );
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document.write( "c:  P( only one machine is working) = 3 * 0.60 * 0.40^2 = 0.288<BR>\n" );
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