document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1160096>Question 1160096</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question3: An insurance salesman sells policies to 5 men, all of identical age. According to the actuarial tables, the probability that a man of this particular age will be alive 30 years hence is 2/3. Find the probability that in 30 years a. All 5 men b. At least 3 men c. Only 2 men d. At least 1 man will be alive. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #783257 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=783257\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a.  This is (2/3)^5=32/243 or 0.132\r<BR>\n" );
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document.write( "Look at the rest of the possibilities:<BR>\n" );
document.write( "For 1 to be alive is 5(2/3)(1/3)^4=10/243<BR>\n" );
document.write( "For 2 to be alive is 10(2/3)^2(1/3)^3=40/243<BR>\n" );
document.write( "for 3, it is 10(2/3)^3(1/3)^2=80/243<BR>\n" );
document.write( "for 4, it is 5(2/3)^4(1/3)=80/243<BR>\n" );
document.write( "For 0, it is (1/3)^5=1/243\r<BR>\n" );
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document.write( "Those all add up to 243/243\r<BR>\n" );
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document.write( "b. At least 3 men means 3,4,5 or 192/243 or 0.7901<BR>\n" );
document.write( "c. For 2 men, it is 40/243 or 0.1646<BR>\n" );
document.write( "d. At least 1 man means 1-p(0 men) or 242/243 or 0.9959\r<BR>\n" );
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document.write( " </SPAN>\n" );
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