document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164034>Question 1164034</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A box contains one yellow, two red, and three green balls. Two balls are randomly chosen without replacement. Define the following events:<BR>\n" );
document.write( "A: { One of the balls is yellow }<BR>\n" );
document.write( "B: { At least one ball is red }<BR>\n" );
document.write( "C: { Both balls are green }<BR>\n" );
document.write( "D: { Both balls are of the same color}\r<BR>\n" );
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document.write( "Find the following conditional probabilities:\r<BR>\n" );
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document.write( "(a)  P(A|B) = \r<BR>\n" );
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document.write( "(b)  P(B^c|D) = \r<BR>\n" );
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document.write( "(c)  P(D|C^c}) = <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854356 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2264)</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=854356\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ```python?code_reference&code_event_index=1<BR>\n" );
document.write( "from itertools import combinations\r<BR>\n" );
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document.write( "# Box contents: 1 Yellow (Y), 2 Red (R1, R2), 3 Green (G1, G2, G3)<BR>\n" );
document.write( "balls = ['Y', 'R1', 'R2', 'G1', 'G2', 'G3']\r<BR>\n" );
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document.write( "# Sample space: Choosing 2 balls without replacement<BR>\n" );
document.write( "S = list(combinations(balls, 2))<BR>\n" );
document.write( "n_S = len(S)\r<BR>\n" );
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document.write( "def is_yellow(combo):<BR>\n" );
document.write( "    return any('Y' in b for b in combo)\r<BR>\n" );
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document.write( "def is_red(combo):<BR>\n" );
document.write( "    return any('R' in b for b in combo)\r<BR>\n" );
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document.write( "def is_green(combo):<BR>\n" );
document.write( "    return all('G' in b for b in combo)\r<BR>\n" );
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document.write( "def same_color(combo):<BR>\n" );
document.write( "    b1, b2 = combo[0][0], combo[1][0]<BR>\n" );
document.write( "    return b1 == b2\r<BR>\n" );
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document.write( "# Events<BR>\n" );
document.write( "A = [c for c in S if is_yellow(c)]<BR>\n" );
document.write( "B = [c for c in S if is_red(c)]<BR>\n" );
document.write( "C = [c for c in S if is_green(c)]<BR>\n" );
document.write( "D = [c for c in S if same_color(c)]\r<BR>\n" );
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document.write( "# (a) P(A|B) = P(A and B) / P(B)<BR>\n" );
document.write( "A_and_B = [c for c in A if c in B]<BR>\n" );
document.write( "p_a_given_b = len(A_and_B) / len(B)\r<BR>\n" );
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document.write( "# (b) P(B^c | D) = P(B^c and D) / P(D)<BR>\n" );
document.write( "not_B = [c for c in S if c not in B]<BR>\n" );
document.write( "not_B_and_D = [c for c in not_B if c in D]<BR>\n" );
document.write( "p_not_b_given_d = len(not_B_and_D) / len(D)\r<BR>\n" );
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document.write( "# (c) P(D | C^c) = P(D and C^c) / P(C^c)<BR>\n" );
document.write( "not_C = [c for c in S if c not in C]<BR>\n" );
document.write( "D_and_not_C = [c for c in D if c in not_C]<BR>\n" );
document.write( "p_d_given_not_c = len(D_and_not_C) / len(not_C)\r<BR>\n" );
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document.write( "print(f\"n(S): {n_S}\")<BR>\n" );
document.write( "print(f\"n(A): {len(A)}, n(B): {len(B)}, n(C): {len(C)}, n(D): {len(D)}\")<BR>\n" );
document.write( "print(f\"n(A and B): {len(A_and_B)}\")<BR>\n" );
document.write( "print(f\"P(A|B): {len(A_and_B)}/{len(B)}\")<BR>\n" );
document.write( "print(f\"P(B^c|D): {len(not_B_and_D)}/{len(D)}\")<BR>\n" );
document.write( "print(f\"P(D|C^c): {len(D_and_not_C)}/{len(not_C)}\")\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "```text?code_stdout&code_event_index=1<BR>\n" );
document.write( "n(S): 15<BR>\n" );
document.write( "n(A): 5, n(B): 9, n(C): 3, n(D): 4<BR>\n" );
document.write( "n(A and B): 2<BR>\n" );
document.write( "P(A|B): 2/9<BR>\n" );
document.write( "P(B^c|D): 3/4<BR>\n" );
document.write( "P(D|C^c): 1/12\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "To solve these conditional probabilities, we first determine the total number of outcomes in the sample space and the number of outcomes for each event.\r<BR>\n" );
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document.write( "The box contains **6 balls**: 1 Yellow ($Y$), 2 Red ($R$), and 3 Green ($G$).<BR>\n" );
document.write( "The total number of ways to choose 2 balls without replacement is:<BR>\n" );
document.write( "$$n(S) = \binom{6}{2} = 15$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Step 1: Define the Event Counts**<BR>\n" );
document.write( "* **Event A (One ball is Yellow):** This means choosing the 1 Yellow ball and 1 from the other 5 balls.<BR>\n" );
document.write( "  $n(A) = \binom{1}{1} \times \binom{5}{1} = 5$<BR>\n" );
document.write( "* **Event B (At least one Red):** Total outcomes minus outcomes with no Red.<BR>\n" );
document.write( "  $n(B) = 15 - \binom{4}{2} = 15 - 6 = 9$<BR>\n" );
document.write( "* **Event C (Both Green):** Choosing 2 from the 3 Green balls.<BR>\n" );
document.write( "  $n(C) = \binom{3}{2} = 3$<BR>\n" );
document.write( "* **Event D (Same Color):** Both Red or Both Green. (Yellow only has one ball, so \"Both Yellow\" is impossible).<BR>\n" );
document.write( "  $n(D) = \binom{2}{2} (\text{Red}) + \binom{3}{2} (\text{Green}) = 1 + 3 = 4$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Step 2: Calculate the Conditional Probabilities**\r<BR>\n" );
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document.write( "**(a) $P(A \mid B)$**<BR>\n" );
document.write( "We need $n(A \cap B)$, which is the number of ways to have 1 Yellow ball **and** at least one Red ball. Since we only pick 2 balls, this must be exactly 1 Yellow and 1 Red.<BR>\n" );
document.write( "$n(A \cap B) = \binom{1}{1} \times \binom{2}{1} = 2$<BR>\n" );
document.write( "$$P(A \mid B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{9}$$\r<BR>\n" );
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document.write( "**(b) $P(B^c \mid D)$**<BR>\n" );
document.write( "$B^c$ is the event \"No Red balls.\" We need $n(B^c \cap D)$, which is the number of ways the balls are the same color **and** neither is Red. This leaves only the \"Both Green\" option.<BR>\n" );
document.write( "$n(B^c \cap D) = \binom{3}{2} = 3$<BR>\n" );
document.write( "$$P(B^c \mid D) = \frac{n(B^c \cap D)}{n(D)} = \frac{3}{4}$$\r<BR>\n" );
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document.write( "**(c) $P(D \mid C^c)$**<BR>\n" );
document.write( "$C^c$ is the event \"Not both Green.\" <BR>\n" );
document.write( "$n(C^c) = n(S) - n(C) = 15 - 3 = 12$<BR>\n" );
document.write( "$n(D \cap C^c)$ is the number of same-color pairs that are **not** Green. This leaves only the \"Both Red\" option.<BR>\n" );
document.write( "$n(D \cap C^c) = \binom{2}{2} = 1$<BR>\n" );
document.write( "$$P(D \mid C^c) = \frac{n(D \cap C^c)}{n(C^c)} = \frac{1}{12}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Final Results:**<BR>\n" );
document.write( "* (a) $P(A \mid B) = \mathbf{2/9}$<BR>\n" );
document.write( "* (b) $P(B^c \mid D) = \mathbf{3/4}$<BR>\n" );
document.write( "* (c) $P(D \mid C^c) = \mathbf{1/12}$ </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );