document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=872700>Question 872700</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Out of each batch of 15 items, 4 are faulty. Items are examined one by one, and items checked are not replaced. <BR>\n" );
document.write( "a) what is the probability that there are exactly 3 defectives in the first 8 examined<BR>\n" );
document.write( "b) probability that the 9th item examined is the 4th defective one found\r<BR>\n" );
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document.write( "I can calculate probability that first item being defective is 4 in 15, and then assuming a faulty item is chosen, the second item has a probability of 3 in 14, or 4 in 14 if the first item selected works. I am sure that I could perform very many multiple calculations and maybe get the correct result, but I am sure there must be a single equation to calculate these that I have missed. I can't find it in my books, and I cannot identify the 'type' of equation to Google examples.<BR>\n" );
document.write( "Any help would be very much appreciated, as my son is trying to complete an IB past paper and between us we are stuck on this one.<BR>\n" );
document.write( "Gary Lynk </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #526353 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=526353\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Out of each batch of 15 items, 4 are faulty. Items are examined one by one, and items checked are not replaced. <BR>\n" );
document.write( "a) what is the probability that there are exactly 3 defectives in the first 8 examined<BR>\n" );
document.write( "This is a Binomial Problem with n = 8 and p(defect) = 4/15<BR>\n" );
document.write( "----<BR>\n" );
document.write( "There are 8C3 = 56 ways to have 3 of 8 defective items<BR>\n" );
document.write( "The probability of eachset of 3 defect and 5 not defect i (4/15)^3*(11/15)^8<BR>\n" );
document.write( "----<BR>\n" );
document.write( "Answer to your Problem::<BR>\n" );
document.write( "P(x = 3 defect in 8) = 56(4/15)^3*(11/15)^8 = 0.2252<BR>\n" );
document.write( "If you use a TI calculator you get binompdf(8,4/15,3) = 0.2252<BR>\n" );
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document.write( "b) probability that the 9th item examined is the 4th defective one found<BR>\n" );
document.write( "Patterm:: nnnnnnnd = (11/15)^7*(4/15) = 0.000000017168..<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "==================== <BR>\n" );
document.write( " </SPAN>\n" );
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