document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1084276>Question 1084276</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if a new drug has been found to be effective 40% of the time, then what is the probability that in random sample of 4 patients ,it will be effective on 2 of them? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #698367 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=698367\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> p(x) = p^x * q^(n-x) * c(n,x)\r<BR>\n" );
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document.write( "when n = 4, x will range from 0 to 4\r<BR>\n" );
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document.write( "p(x) is the probability that x events will occur.\r<BR>\n" );
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document.write( "p is the probability that one event will occur.<BR>\n" );
document.write( "q is the probability that one event will not occur.\r<BR>\n" );
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document.write( "p^x is p raised to the power of x.<BR>\n" );
document.write( "q^(n-x) is q raised to the power of (n-x)\r<BR>\n" );
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document.write( "q is equal to 1 - p.\r<BR>\n" );
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document.write( "c(n,x) is the combination formula for obtaining the number of sets of x elements each out of the pool of n elements when order is not important.\r<BR>\n" );
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document.write( "the formula for c(n,x) is c(n,x) = n! / (x! * (n-x)!)\r<BR>\n" );
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document.write( "in your problem:\r<BR>\n" );
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document.write( "n is equal to 4<BR>\n" );
document.write( "x goes from 0 to 4<BR>\n" );
document.write( "p is equal to .4<BR>\n" );
document.write( "q is equal to .6\r<BR>\n" );
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document.write( "the questions is what is the probability that the drug will be effective on 2 out of 4 patients.\r<BR>\n" );
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document.write( "that would be p(2) = .4^2 * .6^2 * c(4,2)\r<BR>\n" );
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document.write( "that becomes p(2) = .16 * .36 * 6\r<BR>\n" );
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document.write( "the result is p(2) = .3456\r<BR>\n" );
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document.write( "that's your solution.\r<BR>\n" );
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document.write( "the sum of all probabilities has to be equal to 1.\r<BR>\n" );
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document.write( "in this problem, that means that  p(0) + p(1) + p(2) + p(3) + p(4) must be equal to 1.\r<BR>\n" );
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document.write( "i did a check through the use of excel and found that this is true.\r<BR>\n" );
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document.write( "if you find p(0) through p(4) and add them up, they will be equl to 1.\r<BR>\n" );
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document.write( "these are the numbers:\r<BR>\n" );
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document.write( "x       p(x)\r\n" );
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document.write( "0	0.1296\r\n" );
document.write( "1	0.3456\r\n" );
document.write( "2	0.3456\r\n" );
document.write( "3	0.1536\r\n" );
document.write( "4	0.0256\r\n" );
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document.write( "sum     1.0000\r\n" );
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document.write( "p(1) and p(2) are the same probability, but that's not an error.\r<BR>\n" );
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document.write( "p(1) = .4^1 * .6^3 * 4! / (1! * 3!) = .4 * .216 * (4*3*2*1) / (1*3*2*1) = .4 * .216 * 4 = .3456\r<BR>\n" );
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document.write( "p(2) = .4^2 * .6^2 * 4! / (2! * 2!) = .16 * .36 * (4*3*2*1) / (2*1*2*1) = .16 * .36 * 6 = .3456\r<BR>\n" );
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document.write( "the overall probability of each is the same, but the elements within each overall probability are different.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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