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

Question

PQ (q) = { 1/6, if q 1/4, 2/6, ifq2/4, 3/6, ifq3/4, 0, otherwise.

Let K be the total number of Heads in two independent tosses of the coin. Find PQ given K (3/4 given 2)

Answer 1

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