document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=901356>Question 901356</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question 1 - Consider two urns. Urn 1 contains 4 red balls and 2 white balls, and urn 2 contains 3 balls of each color. If 2 balls are drawn from urn 1 without replacement and transferred to urn 2 and then a ball is drawn from urn 2, what is the probability that the ball drawn from urn 2 will be red . Given that the ball drawn from urn 2 was red, what is the conditional probability that (a) 0 (b) 1 (c) 2 red balls are transfered\r<BR>\n" );
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document.write( "Question 2- Let urn 1 contains 4 red balls and 2 white balls, and let the urn 2 contain 3 balls of each color. If a ball is drawn at random from urn 1 and transferred to urn 2 and then a ball is drawn at random from urn 2, what is the probability that the second ball drawn will be red </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #644081 by </B><A HREF=/tutors/aboutme.mpl?userid=Natolino1983><B>Natolino1983(23)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Natolino1983><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=644081\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Q1) Case 1: 2 Red Balls are transfered from Urn1 to Urn 2.<BR>\n" );
document.write( "   Case 2: 1 Red Ball and 1 White Ball are transfered from Urn1 to Urn 2.<BR>\n" );
document.write( "   Case 3: 2 Red White Balls are transfered from Urn1 to Urn 2.\r<BR>\n" );
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document.write( "P(Case 1)= P (First Draw Red & Second Draw Red)=P(First Draw Red) *(Second Draw Red/First Draw Red) = 4/6 * 3/5 = 2/5.\r<BR>\n" );
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document.write( "P(Case 2)= P ((First Draw Red & Second Draw white) or (First draw White & Second draw Red))= P(First Draw red)*P(Second Draw White/First Draw Red) + P(First Draw White)*P(Second Draw Red/First Draw White) =4/6 * 2/5 + 2/6 * 4/5 = 4/15 + 4/15 = 8/15.\r<BR>\n" );
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document.write( "P(Case 3)= P (First Draw White & Second Draw White)=P(First Draw White) *(Second Draw White/First Draw White) = 2/6 * 1/5 = 1/15.\r<BR>\n" );
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document.write( "Note That P(Case 1)+ P(Case 2) + P(Case 3) = 1. (obviously).\r<BR>\n" );
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document.write( "So, 1)P(Draw Red Urn 2) = P(Draw Red Urn 2/Case 1) * (P(Case 1) + P(Draw Red Urn 2/Case 2) * (P(Case 2) + P(Draw Red Urn 2/Case 3) * (P(Case 3) = 5/8 * 2/5 + 4/8 * 8/15 + 3/8 * 1/15 = 1/4 + 4/16 + 1/40 = 21/40.<BR>\n" );
document.write( "    2)a)P(Case 3/Draw Red Urn 2) = (P(Draw Red Urn 2 /Case 3)* (P(Case 3))/ P(Draw Red Urn 2) = (1/40)/(21/40) =1/21<BR>\n" );
document.write( "    2 b) P(Case 2/Draw Red Urn 2) = (P(Draw Red Urn 2 /Case 2)* (P(Case 2))/ P(Draw Red Urn 2) = (4/16)/(21/40) =10/21.<BR>\n" );
document.write( "    2 c)  P(Case 1/Draw Red Urn 2) = 1 - P(Case 3/Draw Red Urn 2) - P(Case 2/Draw Red Urn 2) = 1 - 1/21 -10/21 =10/21.\r<BR>\n" );
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document.write( "You can use Question 1 to solve question 2.<BR>\n" );
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