document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=664071>Question 664071</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am a salesperson & I close on 70% of all my sales.  If I call on 4 sales, what is the probability that I will close on 3 of the 4 sales? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #413104 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=413104\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the binomial probability distribution formula P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "In this case, k = 3, n = 4 and p = 0.7\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)\r<BR>\n" );
document.write( "\n" );
document.write( "P(X = 3) = (4 C 3)*(0.7)^(3)*(1-0.7)^(4-3)\r<BR>\n" );
document.write( "\n" );
document.write( "P(X = 3) = (4 C 3)*(0.7)^(3)*(0.3)^(4-3)\r<BR>\n" );
document.write( "\n" );
document.write( "P(X = 3) = (4)*(0.7)^(3)*(0.3)^1\r<BR>\n" );
document.write( "\n" );
document.write( "P(X = 3) = (4)*(0.343)*(0.3)\r<BR>\n" );
document.write( "\n" );
document.write( "P(X = 3) = 0.4116\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So the probability is 0.4116, which is a 41.16% chance. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );