There are two coins, one unbiased with probability `1/2` or getting heads and the other one is biased with probability `3/4` of getting heads. A coin is selected at random and tossed. It shows heads up. Then the probability that the unbiased coin was selected is

Options

A. `2/3`
B. `3/5`
C. `1/2`
D. `2/5`

Answer

To solve the problem, we will use Bayes' theorem. We need to find the probability that the unbiased coin was selected given that we tossed a head.

### Step-by-step Solution:

1. **Define Events**:
- Let \( E_1 \) be the event of selecting the biased coin.
- Let \( E_2 \) be the event of selecting the unbiased coin.
- Let \( H \) be the event of getting heads.

2. **Determine Probabilities**:
- The probability of selecting either coin is:
\[ P(E_1) = P(E_2) = \frac{1}{2} \]
- The probability of getting heads given the biased coin is selected:
\[ P(H | E_1) = \frac{3}{4} \]
- The probability of getting heads given the unbiased coin is selected:
\[ P(H | E_2) = \frac{1}{2} \]

3. **Calculate Total Probability of Getting Heads**:
- Using the law of total probability:
\[ P(H) = P(H | E_1) \cdot P(E_1) + P(H | E_2) \cdot P(E_2) \]
- Substituting the values:
\[ P(H) = \left(\frac{3}{4} \cdot \frac{1}{2}\right) + \left(\frac{1}{2} \cdot \frac{1}{2}\right) \]
- Simplifying:
\[ P(H) = \frac{3}{8} + \frac{1}{4} = \frac{3}{8} + \frac{2}{8} = \frac{5}{8} \]

4. **Apply Bayes' Theorem**:
- We want to find \( P(E_2 | H) \):
\[ P(E_2 | H) = \frac{P(H | E_2) \cdot P(E_2)}{P(H)} \]
- Substituting the known values:
\[ P(E_2 | H) = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)}{\frac{5}{8}} \]
- Simplifying:
\[ P(E_2 | H) = \frac{\frac{1}{4}}{\frac{5}{8}} = \frac{1}{4} \cdot \frac{8}{5} = \frac{2}{5} \]

5. **Final Answer**:
- The probability that the unbiased coin was selected given that we observed heads is:
\[ \boxed{\frac{2}{5}} \]