document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166412>Question 1166412</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Three bags identical in appearance contain respectively 1 white and 3 red balls ; 2 white and 1 red balls ; 4 white and 3 red balls. One bag is selected randomly and two balls are taken from it. Find the probability that (1) One ball is white and another ball is red (2) A white ball is selected from bag, I, II, III and second ball is red. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852758 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2030)</B></A><img src=\"/images/stars/stars-13.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=852758\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a problem involving **conditional probability** and the **Law of Total Probability**. Since the bags are identical in appearance, the probability of selecting any one bag is $\frac{1}{3}$.\r<BR>\n" );
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document.write( "Let $B_I$, $B_{II}$, and $B_{III}$ be the events that Bag I, Bag II, and Bag III are selected, respectively.<BR>\n" );
document.write( "$$P(B_I) = P(B_{II}) = P(B_{III}) = \frac{1}{3}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## 🎒 Bag Contents\r<BR>\n" );
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document.write( "| Bag | White Balls ($W$) | Red Balls ($R$) | Total Balls ($T$) |<BR>\n" );
document.write( "| :---: | :---: | :---: | :---: |<BR>\n" );
document.write( "| **I** ($B_I$) | 1 | 3 | 4 |<BR>\n" );
document.write( "| **II** ($B_{II}$) | 2 | 1 | 3 |<BR>\n" );
document.write( "| **III** ($B_{III}$) | 4 | 3 | 7 |\r<BR>\n" );
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document.write( "Let $E$ be the event that \"One ball is **White** and the other is **Red**\" (i.e., one $W$ and one $R$) when two balls are drawn without replacement.\r<BR>\n" );
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document.write( "The probability of drawing one white and one red ball from a bag is given by:<BR>\n" );
document.write( "$$P(E|B_i) = \frac{\text{(Ways to choose 1 W)} \times \text{(Ways to choose 1 R)}}{\text{Ways to choose 2 balls from Total}} = \frac{\binom{W}{1} \times \binom{R}{1}}{\binom{T}{2}}$$\r<BR>\n" );
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document.write( "### 1. Calculate Conditional Probabilities $P(E|B_i)$\r<BR>\n" );
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document.write( "* **From Bag I ($B_I$):**<BR>\n" );
document.write( "    $$P(E|B_I) = \frac{\binom{1}{1} \times \binom{3}{1}}{\binom{4}{2}} = \frac{1 \times 3}{6} = \frac{3}{6} = \frac{1}{2}$$\r<BR>\n" );
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document.write( "* **From Bag II ($B_{II}$):**<BR>\n" );
document.write( "    $$P(E|B_{II}) = \frac{\binom{2}{1} \times \binom{1}{1}}{\binom{3}{2}} = \frac{2 \times 1}{3} = \frac{2}{3}$$\r<BR>\n" );
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document.write( "* **From Bag III ($B_{III}$):**<BR>\n" );
document.write( "    $$P(E|B_{III}) = \frac{\binom{4}{1} \times \binom{3}{1}}{\binom{7}{2}} = \frac{4 \times 3}{21} = \frac{12}{21} = \frac{4}{7}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## (1) Probability that One Ball is White and Another Ball is Red\r<BR>\n" );
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document.write( "We use the **Law of Total Probability**:<BR>\n" );
document.write( "$$P(E) = P(E|B_I)P(B_I) + P(E|B_{II})P(B_{II}) + P(E|B_{III})P(B_{III})$$\r<BR>\n" );
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document.write( "$$P(E) = \left(\frac{1}{2} \times \frac{1}{3}\right) + \left(\frac{2}{3} \times \frac{1}{3}\right) + \left(\frac{4}{7} \times \frac{1}{3}\right)$$<BR>\n" );
document.write( "$$P(E) = \frac{1}{3} \left(\frac{1}{2} + \frac{2}{3} + \frac{4}{7}\right)$$\r<BR>\n" );
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document.write( "Find a common denominator for $\frac{1}{2} + \frac{2}{3} + \frac{4}{7}$. The Least Common Multiple (LCM) of 2, 3, and 7 is 42.\r<BR>\n" );
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document.write( "$$P(E) = \frac{1}{3} \left(\frac{21}{42} + \frac{28}{42} + \frac{24}{42}\right)$$<BR>\n" );
document.write( "$$P(E) = \frac{1}{3} \left(\frac{21 + 28 + 24}{42}\right)$$<BR>\n" );
document.write( "$$P(E) = \frac{1}{3} \times \frac{73}{42} = \frac{73}{126}$$\r<BR>\n" );
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document.write( "The probability that one ball is white and the other is red is $\mathbf{\frac{73}{126}}$.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## (2) A white ball is selected from bag, I, II, III and second ball is red\r<BR>\n" );
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document.write( "This part asks for the probability of the **compound event** that a specific bag is chosen **AND** the draw results in one white and one red ball.\r<BR>\n" );
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document.write( "Let $E$ be the event that \"One ball is White and one ball is Red.\"<BR>\n" );
document.write( "The prompt requires the sum of the probabilities of the three mutually exclusive events: $(E \text{ and } B_I)$, $(E \text{ and } B_{II})$, and $(E \text{ and } B_{III})$.\r<BR>\n" );
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document.write( "$$P(\text{Compound Events}) = P(E \cap B_I) + P(E \cap B_{II}) + P(E \cap B_{III})$$\r<BR>\n" );
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document.write( "Using the multiplication rule, $P(E \cap B_i) = P(E|B_i) \cdot P(B_i)$:\r<BR>\n" );
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document.write( "$$P(\text{Sum}) = \left(\frac{1}{2} \times \frac{1}{3}\right) + \left(\frac{2}{3} \times \frac{1}{3}\right) + \left(\frac{4}{7} \times \frac{1}{3}\right)$$\r<BR>\n" );
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document.write( "This is exactly the same calculation as Part (1), which is the total probability $P(E)$.\r<BR>\n" );
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document.write( "$$P(\text{Sum}) = \frac{1}{6} + \frac{2}{9} + \frac{4}{21}$$<BR>\n" );
document.write( "$$P(\text{Sum}) = \frac{73}{126}$$\r<BR>\n" );
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document.write( "The probability that the draw results in one white and one red ball, summing across all three bags, is $\mathbf{\frac{73}{126}}$. </SPAN>\n" );
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