document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1167372>Question 1167372</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Assume that when adults with smartphones are randomly​ selected, 55​% use them in meetings or classes. If 7  adult smartphone users are randomly​ selected, find the probability that at least 4 of them use their smartphones in meetings or classes. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #791973 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=791973\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a binomial with n=7 p=0.55 <BR>\n" );
document.write( "a least 4 of them use their smartphones<BR>\n" );
document.write( "that is 1-prob(3) of them use them.<BR>\n" );
document.write( "=0.6083 is answer.<BR>\n" );
document.write( "probability 4 use them is 7C4*0.55^4*0.45^3=0.2918 <BR>\n" );
document.write( "for 5 it is 0.2140<BR>\n" );
document.write( "for 6 it is 0.0872<BR>\n" );
document.write( "for 7 it is 0.55^7 or 0.0152<BR>\n" );
document.write( "add them to get 0.6082 (rounding error)<BR>\n" );
document.write( " </SPAN>\n" );
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