document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=479505>Question 479505</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am having difficulties solving the following problem:<BR>\n" );
document.write( "Nine tickets, numbered 1 to 9, are in a box.  If 2 tickets are drawn, determine the probability p that one is odd and one is even.  \r<BR>\n" );
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document.write( "Here is what I did (Method 1): <BR>\n" );
document.write( "P(1=even, 9 tickets)<BR>\n" );
document.write( "n (total) = 9, n (even) = 5, n (odd) = 4<BR>\n" );
document.write( "P(1=odd, 9 tickets) <BR>\n" );
document.write( "P (A and B) = P(A) times P(B)<BR>\n" );
document.write( "[(1/9)X4]X[(1/8)X5] = 5/18 However, my textbook's answer is 5/9. <BR>\n" );
document.write( "Even if I draw even ticket first, then odd ticket, I will get 5/9 since:<BR>\n" );
document.write( "[(1/9)X5]X[(1/8)times4] = 5/18. \r<BR>\n" );
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document.write( "Method 2:<BR>\n" );
document.write( "5C1/9C1 = 5/9<BR>\n" );
document.write( "4C1/8C1 = 1/2<BR>\n" );
document.write( "P(A and B) = 5/9 X 1/2 = 5/18. While it confirms my answer obtained through Method 1, the textbook's answer is 5/9. \r<BR>\n" );
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document.write( "Prior solving this problem, I thought about it this way: \r<BR>\n" );
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document.write( "We have 9 total tickets, 5 are even, and 4 are odd. <BR>\n" );
document.write( "Since we are dealing with tickets, common sense tells me 1. we are taking one ticket at a time; 2. tickets cannot be replaced; 3. after I draw my first ticket, I will have one less ticket and will draw my second ticket from smaller number of tickets, thus we have dependant events; 4. combinations is one way to solve this problem; 5. multiply probability of Event 1 with probability of Event 2 to get total probability.\r<BR>\n" );
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document.write( "Here is how the textbook did it: <BR>\n" );
document.write( "(5C1 X 4C1) / 9C2 = 5/9. <BR>\n" );
document.write( "Please note how they are drawing first ticket and second ticket from 9 tickets as if cards WERE REPLACED. WHY? \r<BR>\n" );
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document.write( "Please let me know where I went wrong in my reasoning. \r<BR>\n" );
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document.write( "Thank you very much.   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #328530 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=328530\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Nine tickets, numbered 1 to 9, are in a box. If 2 tickets are drawn, determine the probability p that one is odd and one is even. <BR>\n" );
document.write( "Here is what I did (Method 1): <BR>\n" );
document.write( "P(1=even, 9 tickets)<BR>\n" );
document.write( "n (total) = 9, n (even) = 5, n (odd) = 4<BR>\n" );
document.write( "P(1=odd, 9 tickets) <BR>\n" );
document.write( "P (A and B) = P(A) times P(B)<BR>\n" );
document.write( "[(1/9)X4]X[(1/8)X5] = 5/18 However, my textbook's answer is 5/9. <BR>\n" );
document.write( "Even if I draw even ticket first, then odd ticket, I will get 5/9 since:<BR>\n" );
document.write( "[(1/9)X5]X[(1/8)times4] = 5/18. <BR>\n" );
document.write( "Method 2:<BR>\n" );
document.write( "5C1/9C1 = 5/9<BR>\n" );
document.write( "4C1/8C1 = 1/2<BR>\n" );
document.write( "P(A and B) = 5/9 X 1/2 = 5/18. While it confirms my answer obtained through Method 1, the textbook's answer is 5/9. <BR>\n" );
document.write( "Prior solving this problem, I thought about it this way: <BR>\n" );
document.write( "We have 9 total tickets, 5 are even, and 4 are odd. <BR>\n" );
document.write( "Since we are dealing with tickets, common sense tells me 1. we are taking one ticket at a time; 2. tickets cannot be replaced; 3. after I draw my first ticket, I will have one less ticket and will draw my second ticket from smaller number of tickets, thus we have dependant events; 4. combinations is one way to solve this problem; 5. multiply probability of Event 1 with probability of Event 2 to get total probability. <BR>\n" );
document.write( "Here is how the textbook did it: <BR>\n" );
document.write( "(5C1 X 4C1) / 9C2 = 5/9. <BR>\n" );
document.write( "Please note how they are drawing first ticket and second ticket from 9 tickets as if cards WERE REPLACED. WHY? <BR>\n" );
document.write( "Please let me know where I went wrong in my reasoning. <BR>\n" );
document.write( "======================================================<BR>\n" );
document.write( "Nine tickets, numbered 1 to 9, are in a box. If 2 tickets are drawn, determine the probability p that one is odd and one is even. <BR>\n" );
document.write( "---<BR>\n" );
document.write( "There are 3 possible results: 2 odd ; 2 even ; 1 odd and 1 even<BR>\n" );
document.write( "---<BR>\n" );
document.write( "# of even digits = 4<BR>\n" );
document.write( "# of odd digits = 5<BR>\n" );
document.write( "----<BR>\n" );
document.write( "P(2 even) = 4C2/9C2 = 6/36 = 1/6<BR>\n" );
document.write( "P(2 odd) = 5C2/9C2 = 10/36 = 5/18<BR>\n" );
document.write( "----<BR>\n" );
document.write( "P(one odd and one even) <BR>\n" );
document.write( "= 1 - [P(2 even)+P(2 odd)] <BR>\n" );
document.write( "= 1-[(3/18)+5(18)] <BR>\n" );
document.write( "= 10/18<BR>\n" );
document.write( "= 5/9<BR>\n" );
document.write( "==============<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "============== </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );