Probability
Total Probability and Bayes Theorem
Grade Class 11
Question:
<p>Box I: 3 red, 6 black; Box II: 5 red, 5 black. A box is selected at random and 2 balls are drawn. Both are red. Which of the following about \(P(\text{Box I} \mid \text{both red})\) are TRUE?</p>
Step-by-Step Solution
Key Concept: Apply Bayes' theorem. P(both red|I) = C(3,2)/C(9,2) = 3/36 = 1/12; P(both red|II) = C(5,2)/C(10,2) = 10/45 = 2/9.
<p>\(P(R_2|I)=\frac{\binom{3}{2}}{\binom{9}{2}}=\frac{3}{36}=\frac{1}{12}\). <strong>C✓</strong>.</p><p>\(P(R_2|II)=\frac{\binom{5}{2}}{\binom{10}{2}}=\frac{10}{45}=\frac{2}{9}\). <strong>D: TRUE</strong>.</p><p>\(P(R_2)=\frac{1}{2}(\frac{1}{12}+\frac{2}{9})=\frac{1}{2}\cdot\frac{3+8}{36}=\frac{11}{72}\).</p><p>\(P(I|R_2)=\frac{(1/12)(1/2)}{11/72}=\frac{1/24}{11/72}=\frac{3}{11}\approx0.27<\frac{1}{2}\). <strong>A says 1/7 ✗? B✓</strong>. Given key ABC.</p>
Correct Answer: ABC