document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1184146>Question 1184146</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A business office orders paper supplies from one of three vendors, V1,<BR>\n" );
document.write( "V2, or V3. Orders are to be placed on two successive days, one order<BR>\n" );
document.write( "per day. Thus, (V2,V3) might denote that vendor V2 gets the order on<BR>\n" );
document.write( "the first day and vendor V3 gets the order on the second day.<BR>\n" );
document.write( "a. List the sample points in this experiment of ordering paper on two<BR>\n" );
document.write( "successive days.<BR>\n" );
document.write( "b. Assume the vendors are selected at random each day and assign a<BR>\n" );
document.write( "probability to each sample point.<BR>\n" );
document.write( "c. Let A denote the event that the same vendor gets both orders and B<BR>\n" );
document.write( "the event that V2 gets at least one order. Find P(A), P(B), P(A∪ B)and<BR>\n" );
document.write( "P(A∩ B)by summing the probabilities of the sample points in these<BR>\n" );
document.write( "events </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849874 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=849874\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's the solution:\r<BR>\n" );
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document.write( "**a. Sample Points:**\r<BR>\n" );
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document.write( "The sample points represent all possible pairs of vendors chosen on two successive days.  We can list them as ordered pairs (Vendor Day 1, Vendor Day 2):\r<BR>\n" );
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document.write( "* (V1, V1)<BR>\n" );
document.write( "* (V1, V2)<BR>\n" );
document.write( "* (V1, V3)<BR>\n" );
document.write( "* (V2, V1)<BR>\n" );
document.write( "* (V2, V2)<BR>\n" );
document.write( "* (V2, V3)<BR>\n" );
document.write( "* (V3, V1)<BR>\n" );
document.write( "* (V3, V2)<BR>\n" );
document.write( "* (V3, V3)\r<BR>\n" );
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document.write( "**b. Probabilities:**\r<BR>\n" );
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document.write( "Since the vendors are selected at random each day, each vendor has an equal probability of being chosen (1/3).  Since the orders on the two days are independent events, we can find the probability of each sample point by multiplying the probabilities for each day.\r<BR>\n" );
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document.write( "* P(V1, V1) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V1, V2) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V1, V3) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V2, V1) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V2, V2) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V2, V3) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V3, V1) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V3, V2) = (1/3) * (1/3) = 1/9<BR>\n" );
document.write( "* P(V3, V3) = (1/3) * (1/3) = 1/9\r<BR>\n" );
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document.write( "**c. Probabilities of Events A and B:**\r<BR>\n" );
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document.write( "* **Event A (Same vendor gets both orders):**<BR>\n" );
document.write( "   A = {(V1, V1), (V2, V2), (V3, V3)}<BR>\n" );
document.write( "   P(A) = P(V1, V1) + P(V2, V2) + P(V3, V3) = (1/9) + (1/9) + (1/9) = 3/9 = 1/3\r<BR>\n" );
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document.write( "* **Event B (V2 gets at least one order):**<BR>\n" );
document.write( "   B = {(V1, V2), (V2, V1), (V2, V2), (V2, V3), (V3, V2)}<BR>\n" );
document.write( "   P(B) = P(V1, V2) + P(V2, V1) + P(V2, V2) + P(V2, V3) + P(V3, V2) = (1/9) + (1/9) + (1/9) + (1/9) + (1/9) = 5/9\r<BR>\n" );
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document.write( "* **Event A ∪ B (A or B occurs):**<BR>\n" );
document.write( "   A ∪ B = {(V1, V1), (V2, V2), (V3, V3), (V1, V2), (V2, V1), (V2, V3), (V3, V2)}<BR>\n" );
document.write( "   P(A ∪ B) = P(V1, V1) + P(V2, V2) + P(V3, V3) + P(V1, V2) + P(V2, V1) + P(V2, V3) + P(V3, V2) = (1/9) + (1/9) + (1/9) + (1/9) + (1/9) + (1/9) + (1/9) = 7/9<BR>\n" );
document.write( "   *Alternatively*, since A is a subset of B, P(A∪B) = P(B) = 5/9. I made a mistake in the previous calculation.\r<BR>\n" );
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document.write( "* **Event A ∩ B (A and B occur):**<BR>\n" );
document.write( "   A ∩ B = {(V2, V2)}<BR>\n" );
document.write( "   P(A ∩ B) = P(V2, V2) = 1/9<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );