The bias of a coin (ie, the probability of Heads) can take three possible values, 1/4, 1/2, or 3/4, and is modeled as a discrete random variable Q with PMF

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 \( P(K = 2 | Q = q) \)

1. If \( Q = \frac{1}{4} \):
\[
P(K = 2 | Q = \frac{1}{4}) = P(HH) = \left(\frac{1}{4}\right)^2 = \frac{1}{16}
\]

2. If \( Q = \frac{1}{2} \):
\[
P(K = 2 | Q = \frac{1}{2}) = P(HH) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}
\]

3. 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:
\[
P(Q = \frac{3}{4} | K = 2) = \frac{3}{4}
\]