document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=983415>Question 983415</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. 10 % of article made by a company are defective, if 5 article are selected at random, what is the probability that (i)3 are defective. (ii) less than 2 are defective. (iii) more than 2 are defective.\r<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 #604248 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=604248\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First is 5C3*(0.1)^3(0.9)^2; number of ways 3 can be defective out of 5 times the probability is 0.0081  (0.01*0.81)\r<BR>\n" );
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document.write( "Fewer than 2 means 1 or 0 are defective<BR>\n" );
document.write( "P(1)=5C1(0.1)^1(0.9^4)=0.328<BR>\n" );
document.write( "P(0)=(.9^5)=0.590<BR>\n" );
document.write( "The probability fewer than 2 are defective is 0.918.  That makes sense since it is a low probability to find one defective.\r<BR>\n" );
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document.write( "Probability of 2 defective is 5C2(0.1^2)(0.9^3)=0.073<BR>\n" );
document.write( "The probability of 2 or fewer defective (using the above) is 0.918+0.073=0.991.<BR>\n" );
document.write( "Therefore, the complement is greater than 2 are defective, which is 0.009. </SPAN>\n" );
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