document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1206319>Question 1206319</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A sample is selected from one of two populations, <BR>\n" );
document.write( "S1 and S2,<BR>\n" );
document.write( " with <BR>\n" );
document.write( "P(S1) = 0.7<BR>\n" );
document.write( " and <BR>\n" );
document.write( "P(S2) = 0.3.<BR>\n" );
document.write( " The probabilities that an event A occurs, given that event <BR>\n" );
document.write( "S1<BR>\n" );
document.write( " or <BR>\n" );
document.write( "S2<BR>\n" );
document.write( " has occurred are<BR>\n" );
document.write( "P(A|S1) = 0.2<BR>\n" );
document.write( " and <BR>\n" );
document.write( "P(A|S2) = 0.1.<BR>\n" );
document.write( "Use the Law of Total Probability to find P(A). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #843682 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=843682\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i think this works like this.\r<BR>\n" );
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document.write( "law of total probability states that p(A) = p(A intersect B) plus p(A intersect complement of B.\r<BR>\n" );
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document.write( "you are given:\r<BR>\n" );
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document.write( "p(S1) = .7<BR>\n" );
document.write( "p(S2) = .3\r<BR>\n" );
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document.write( "you are given that p(A given S1) = .2.<BR>\n" );
document.write( "you are given that p(A given S2) = .1\r<BR>\n" );
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document.write( "p(A given S1) is equal to p(A intersect S1) / p(S1).<BR>\n" );
document.write( "this becomes .2 = p(A intersect S1) / .7<BR>\n" );
document.write( "solve for p(A intersect S1) to get:<BR>\n" );
document.write( "p(A intersect S1) = .7 * .2 = .14\r<BR>\n" );
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document.write( "p(A given S2) is equal to p(A intersect S1) / p(S2).<BR>\n" );
document.write( "this becomes .1 = p(A intersect S1) / .3<BR>\n" );
document.write( "solve for p(A intersect S2) to get:<BR>\n" );
document.write( "p(A intersect S2) = .1 * .3 = .03\r<BR>\n" );
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document.write( "p(A) is equal to p(A intersect S1) + p(A intersect S2) = .14 + .03 = .17\r<BR>\n" );
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document.write( "that should be your answer.\r<BR>\n" );
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document.write( "here's a reference.<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://www.statisticshowto.com/total-probability-rule/\" target = \"_blank\">https://www.statisticshowto.com/total-probability-rule/</a>\r<BR>\n" );
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document.write( "i drew a tree diagram (not very pretty) that shows the same result.<BR>\n" );
document.write( "see below.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2024/022901.jpg\">\r<BR>\n" );
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document.write( "in the diagram, that's p(S1 and S2) and p(A and not A).<BR>\n" );
document.write( "the top of the tree is marked as 1, meaning it's 100% probability on top of the tree.<BR>\n" );
document.write( "that encompasses all that's beneath it.<BR>\n" );
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