document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1118588>Question 1118588</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A pharmacist wishes to select three brands of aspirin to sell in his store. He has five major brands to choose from: A, B, C, D and E. If he selects the three brands at random, what is the probability that he will select the following?<BR>\n" );
document.write( "(a)<BR>\n" );
document.write( "brand B<BR>\n" );
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document.write( "(b)<BR>\n" );
document.write( "brands B and C<BR>\n" );
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document.write( "(c)<BR>\n" );
document.write( "at least one of the two brands B and C<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #734129 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=734129\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The number of ways he can choose 3 of the 5 brands is C(5,3) = 10.  So the denominator of the probability fraction is 10.<br><BR>\n" );
document.write( "The numerators are the numbers of ways of choosing the specified brands.<br><BR>\n" );
document.write( "(a) brand B<br><BR>\n" );
document.write( "He has to choose brand B (C(1,1)); and he can choose any 2 of the other 4 (C(4,2)).  The probability is then (C(1,1)*C(4,2))/C(5,3) = (1*6)/10 = 6/10 = 3/5.<br><BR>\n" );
document.write( "(b) brands B and C<br><BR>\n" );
document.write( "He has to choose both brands B and C (C(2,2)); and then he can choose any 1 of the other 3 (C(3,1).  You can figure out the probability for this one.<br><BR>\n" );
document.write( "(c) at least one of the two brands B and C<br><BR>\n" );
document.write( "The opposite of this is choosing neither B nor C and all 3 of the other brands. There is clearly only one way to do that; the probability of choosing neither B nor C is 1/10; so the probability of choosing at least one of B or C is 1-1/10 = 9/10.<br><BR>\n" );
document.write( "Note that the number of different ways to choose 3 of the 5 brands (10) is so small that you can verify any answers you get using formal combinatorical methods by listing all ten combinations and seeing which ones satisfy the given conditions. </SPAN>\n" );
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