To find \( P(Q = \frac{3}{4} | K = 2) \), we can use Bayes' theorem:
\[ P(Q = \frac{3}{4} | K = 2) = \frac{P(K = 2 | Q = \frac{3}{4}) P(Q = \frac{3}{4})}{P(K = 2)} \]
Step 1: Calculate \( P(K = 2 | Q = q) \)
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If \( Q = \frac{1}{4} \): \[ P(K = 2 | Q = \frac{1}{4}) = P(HH) = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
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If \( Q = \frac{1}{2} \): \[ P(K = 2 | Q = \frac{1}{2}) = P(HH) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
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If \( Q = \frac{3}{4} \): \[ P(K = 2 | Q = \frac{3}{4}) = P(HH) = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]
Step 2: Calculate \( P(Q = q) \)
From the problem:
- \( P(Q = \frac{1}{4}) = \frac{1}{6} \)
- \( P(Q = \frac{1}{2}) = \frac{2}{6} = \frac{1}{3} \)
- \( P(Q = \frac{3}{4}) = \frac{3}{6} = \frac{1}{2} \)
Step 3: Calculate \( P(K = 2) \)
Using the law of total probability: \[ P(K = 2) = P(K = 2 | Q = \frac{1}{4}) P(Q = \frac{1}{4}) + P(K = 2 | Q = \frac{1}{2}) P(Q = \frac{1}{2}) + P(K = 2 | Q = \frac{3}{4}) P(Q = \frac{3}{4}) \]
Calculating each term: \[ P(K = 2 | Q = \frac{1}{4}) P(Q = \frac{1}{4}) = \frac{1}{16} \cdot \frac{1}{6} = \frac{1}{96} \]
\[ P(K = 2 | Q = \frac{1}{2}) P(Q = \frac{1}{2}) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12} \]
\[ P(K = 2 | Q = \frac{3}{4}) P(Q = \frac{3}{4}) = \frac{9}{16} \cdot \frac{1}{2} = \frac{9}{32} \]
Now calculate \( P(K = 2) \): \[ P(K = 2) = \frac{1}{96} + \frac{1}{12} + \frac{9}{32} \]
Finding a common denominator: The least common multiple of \( 96, 12, \) and \( 32 \) is \( 192 \).
Converting each term: \[ \frac{1}{96} = \frac{2}{192}, \quad \frac{1}{12} = \frac{16}{192}, \quad \frac{9}{32} = \frac{54}{192} \]
Adding them up: \[ P(K = 2) = \frac{2 + 16 + 54}{192} = \frac{72}{192} = \frac{3}{8} \]
Step 4: Calculate \( P(Q = \frac{3}{4} | K = 2) \)
Now substitute into Bayes' theorem: \[ P(Q = \frac{3}{4} | K = 2) = \frac{P(K = 2 | Q = \frac{3}{4}) P(Q = \frac{3}{4})}{P(K = 2)} = \frac{\frac{9}{16} \cdot \frac{1}{2}}{\frac{3}{8}} \]
Calculating: \[ = \frac{\frac{9}{32}}{\frac{3}{8}} = \frac{9}{32} \cdot \frac{8}{3} = \frac{9 \cdot 8}{32 \cdot 3} = \frac{72}{96} = \frac{3}{4} \]
Thus, the final answer is: \[ P(Q = \frac{3}{4} | K = 2) = \frac{3}{4} \]