document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1177190>Question 1177190</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If P(B) > 0,<BR>\n" );
document.write( "(a) Show that P(A|B) + P(A^c|B) = 1.<BR>\n" );
document.write( "(b) Show that in general the following two statements are false: (i) P(A|B)+ P(A|B^c) = 1,<BR>\n" );
document.write( "(ii) P(A|B) + P(A^c|B^c) = 1\r<BR>\n" );
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document.write( "thank you :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850505 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850505\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's break down this problem step-by-step.\r<BR>\n" );
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document.write( "**Given:**\r<BR>\n" );
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document.write( "* P(B) > 0\r<BR>\n" );
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document.write( "**(a) Show that P(A|B) + P(Aᶜ|B) = 1**\r<BR>\n" );
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document.write( "1.  **Definition of Conditional Probability:**<BR>\n" );
document.write( "    * P(A|B) = P(A ∩ B) / P(B)<BR>\n" );
document.write( "    * P(Aᶜ|B) = P(Aᶜ ∩ B) / P(B)\r<BR>\n" );
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document.write( "2.  **Sum the Conditional Probabilities:**<BR>\n" );
document.write( "    * P(A|B) + P(Aᶜ|B) = [P(A ∩ B) / P(B)] + [P(Aᶜ ∩ B) / P(B)]<BR>\n" );
document.write( "    * P(A|B) + P(Aᶜ|B) = [P(A ∩ B) + P(Aᶜ ∩ B)] / P(B)\r<BR>\n" );
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document.write( "3.  **Recognize the Union:**<BR>\n" );
document.write( "    * Note that (A ∩ B) and (Aᶜ ∩ B) are mutually exclusive events (they cannot occur simultaneously).<BR>\n" );
document.write( "    * Also, their union is (A ∩ B) ∪ (Aᶜ ∩ B) = (A ∪ Aᶜ) ∩ B = S ∩ B = B, where S is the sample space.\r<BR>\n" );
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document.write( "4.  **Simplify the Sum:**<BR>\n" );
document.write( "    * P(A ∩ B) + P(Aᶜ ∩ B) = P(B)\r<BR>\n" );
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document.write( "5.  **Substitute and Conclude:**<BR>\n" );
document.write( "    * P(A|B) + P(Aᶜ|B) = P(B) / P(B) = 1\r<BR>\n" );
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document.write( "**(b) Show that in general the following two statements are false:**\r<BR>\n" );
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document.write( "**(i) P(A|B) + P(A|Bᶜ) = 1**\r<BR>\n" );
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document.write( "1.  **Counterexample:**<BR>\n" );
document.write( "    * Let's consider a simple example.<BR>\n" );
document.write( "    * Suppose we have a fair coin toss.<BR>\n" );
document.write( "    * Let A be the event \"heads\" and B be the event \"the coin is fair.\"<BR>\n" );
document.write( "    * P(A) = 1/2, P(B) = 1.<BR>\n" );
document.write( "    * P(A|B) = 1/2 (probability of heads given the coin is fair).<BR>\n" );
document.write( "    * Bᶜ means \"the coin is not fair,\" which is a rare event.<BR>\n" );
document.write( "    * Let's say P(Bᶜ) = 0.01.<BR>\n" );
document.write( "    * Let's assume P(A|Bᶜ) = 0.8 (if the coin is not fair, it's more likely to be heads).<BR>\n" );
document.write( "    * P(A|B) + P(A|Bᶜ) = 1/2 + 0.8 = 1.3 ≠ 1.\r<BR>\n" );
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document.write( "2.  **General Explanation:**<BR>\n" );
document.write( "    * P(A|B) and P(A|Bᶜ) are conditional probabilities based on different events (B and Bᶜ).<BR>\n" );
document.write( "    * There is no reason to expect that the sum of these probabilities will always be 1.<BR>\n" );
document.write( "    * They are not complementary events.\r<BR>\n" );
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document.write( "**(ii) P(A|B) + P(Aᶜ|Bᶜ) = 1**\r<BR>\n" );
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document.write( "1.  **Counterexample:**<BR>\n" );
document.write( "    * Using the same example as above.<BR>\n" );
document.write( "    * P(A|B) = 1/2<BR>\n" );
document.write( "    * P(Aᶜ|Bᶜ) = P(tails|unfair)<BR>\n" );
document.write( "    * Let's assume P(Aᶜ|Bᶜ) = 0.2 (if the coin is unfair, it's less likely to be tails).<BR>\n" );
document.write( "    * P(A|B) + P(Aᶜ|Bᶜ) = 1/2 + 0.2 = 0.7 ≠ 1.\r<BR>\n" );
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document.write( "2.  **General Explanation:**<BR>\n" );
document.write( "    * P(A|B) and P(Aᶜ|Bᶜ) are conditional probabilities involving different events (B and Bᶜ) and different outcomes (A and Aᶜ).<BR>\n" );
document.write( "    * There is no general relationship that would guarantee their sum to be 1.<BR>\n" );
document.write( "    * These events are not complementary.\r<BR>\n" );
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document.write( "**Conclusion:**\r<BR>\n" );
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document.write( "* **(a)** P(A|B) + P(Aᶜ|B) = 1 is always true when P(B) > 0.<BR>\n" );
document.write( "* **(b)** The statements P(A|B) + P(A|Bᶜ) = 1 and P(A|B) + P(Aᶜ|Bᶜ) = 1 are generally false.<BR>\n" );
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