document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1112221>Question 1112221</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The probability that certain production process will produce a defective parts is .20. Find the exact probability that lot of 12 parts will contain<BR>\n" );
document.write( "a.exactly 5 defective parts<BR>\n" );
document.write( "b.3 or 4 defective parts\r<BR>\n" );
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document.write( "Complete solution please thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #727259 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=727259\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the binomial probability(P) formula<BR>\n" );
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document.write( "P ( k successes in n trials ) = nCk * p^(k) * (1-p)^(n-k), where nCk = n!/(r! * (n-r)!)<BR>\n" );
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document.write( "a) p=0.20, n=12<BR>\n" );
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document.write( "P ( 5 defects in 12 trials ) = 12C5 * (0.20)^5 * (1-0.20)^(12-5) = 0.0532<BR>\n" );
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document.write( "b) P ( 3 defects in 12 trials ) = 12C3 * (0.20)^3 * (1-0.20)^(12-3) = 0.2362<BR>\n" );
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document.write( "P ( 4 defects in 12 trials ) = 12C4 * (0.20)^4 * (1-0.20)^(12-4) = 0.1329<BR>\n" );
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document.write( "P ( 3 or 4 defects in 12 trials ) = 0.2362 + 0.1329 = 0.3691<BR>\n" );
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