document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=192571>Question 192571</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/9780470128633/algebrahomeworkh>Finite Mathematics</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Okay This was a two part question I got the second part correct but cannot seem to get the first part of the question correct.\r<BR>\n" );
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document.write( "Defective Refrigerator Through a mix-up on the production line, 6 defective refrigerators were shipped out with 39 good ones. \r<BR>\n" );
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document.write( "1) If 5 are selected at random, what is the probability that all 5 are defective? Round your answer to 6 decimal places.\r<BR>\n" );
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document.write( "2) What is the probability that at least 2 of them are defective? Round your answer to 3 decimal places. ANS = 0.125 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #144572 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=144572\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "The first part is a conditional probability problem.  Since there are 6 defective refrigerators in a batch of 39, and you select 1 of them, then the probability that you will have selected a defective one is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE  \frac{6}{39}\">.\r<BR>\n" );
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document.write( "Now, given that you were successful and selected a defective unit on the first try, and making the entirely reasonable assumption that, having found a bad one, you wouldn't put it back, when you go to select the second random unit, there are only 5 defective ones remaining out of 38 total. (1 less defective one and 1 less total to choose from).  Hence, the probability on the second selection is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE  \frac{5}{38}\">.\r<BR>\n" );
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document.write( "Continuing in that vein, each time presuming that you were successful on the previous trial, the remaining probabilities are:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{37},\,\frac{3}{36},\,\text{ and }\frac{2}{35}\">\r<BR>\n" );
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document.write( "And finally, the total probability of 5 defective out of 5 selected from 39 is the product of those 5 probabilities, or:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{6}{39}\,\times\,\frac{5}{38}\,\times\,\frac{4}{37}\,\times\,\frac{3}{36}\,\times\,\frac{2}{35}\">\r<BR>\n" );
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document.write( "You can do the arithmetic.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
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