document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=495961>Question 495961</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this problem: \"On a particular production line the probability that an item will be defective is 0.1. A quality controller selects, at random, three items from a batch and examines them for defects. If at least 2 of the items are defective then the whole batch is rejected. Determine the probability that a batch is rejected using the Binomial distribution.\"<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 #336274 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=336274\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> probability of an item being defective is .1<BR>\n" );
document.write( "probability of an item not being defective is 1 - .1 = .9<BR>\n" );
document.write( "if p = defective and q = not defective, then the formula for x out of n would be:<BR>\n" );
document.write( "p(x) = p^x * q^(n-x) * nCx<BR>\n" );
document.write( "we are drawing 3 items from the batch, so n = 3<BR>\n" );
document.write( "we want to find the probability that at least 2 are defective out of the 3.<BR>\n" );
document.write( "this would be equal to 1 - p(0) - p(1) or it would be equal to p(2) + p(3).<BR>\n" );
document.write( "we'll work it both ways so you can see how that happens.<BR>\n" );
document.write( "nCx is the combination formula for x things out of n.<BR>\n" );
document.write( "it is equal to n! / (x! * (n-x)!)<BR>\n" );
document.write( "using the formula of p(x), we get:<BR>\n" );
document.write( "p(0) = .1^0 * .9^3 * 3C0 = .729 * 1 = .729<BR>\n" );
document.write( "p(1) = .1^1 * .9^2 * 3C1 = .081 * 3 = .243<BR>\n" );
document.write( "p(2) = .1^2 * .9^1 * 3C2 = .009 * 3 = .027<BR>\n" );
document.write( "p(3) = .1^3 * .9^0 * 3C3 = .001 * 1 = .001<BR>\n" );
document.write( "sum up all the probabilities and you get p(all) = 1<BR>\n" );
document.write( "this is what you should get when you add up all the probabilities of independent events like we have here.<BR>\n" );
document.write( "p(>=2) = p(2) + p(3) = .027 + .001 = .028<BR>\n" );
document.write( "p(>=2) = 1 - p(0) - p(1) = 1 - .729 - .243 = .028<BR>\n" );
document.write( "we get the same answer either way, as we should.<BR>\n" );
document.write( "your answer is that the probability of getting a sample with at least 2 defective units in it is .028 or 2.8%\r<BR>\n" );
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