There are three coins one is a two-headed coin having head on both faces , another is a biased , coin that coines up tails 25% of the times and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows head what is the probability that it was a two-headed coin ?

Answer

To solve the problem, we will use Bayes' theorem to find the probability that the coin chosen was the two-headed coin (C1) given that the outcome of the toss was a head (H).

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( C_1 \): the event that the chosen coin is the two-headed coin.
- Let \( C_2 \): the event that the chosen coin is the biased coin.
- Let \( C_3 \): the event that the chosen coin is the unbiased coin.
- Let \( H \): the event that the outcome of the toss is a head.

2. **Determine the Prior Probabilities**:
Since one of the three coins is chosen at random, the prior probabilities are:
\[ P(C_1) = P(C_2) = P(C_3) = \frac{1}{3} \]

3. **Determine the Conditional Probabilities**:
- For the two-headed coin \( C_1 \):
\[ P(H | C_1) = 1 \quad \text{(since it always shows heads)} \]
- For the biased coin \( C_2 \) (which shows tails 25% of the time):
\[ P(H | C_2) = 1 - 0.25 = 0.75 = \frac{3}{4} \]
- For the unbiased coin \( C_3 \):
\[ P(H | C_3) = 0.5 = \frac{1}{2} \]

4. **Calculate the Total Probability of Getting Heads**:
Using the law of total probability:
\[ P(H) = P(H | C_1) P(C_1) + P(H | C_2) P(C_2) + P(H | C_3) P(C_3) \]
Substituting the values:
\[ P(H) = 1 \cdot \frac{1}{3} + \frac{3}{4} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} \]
\[ P(H) = \frac{1}{3} + \frac{3}{12} + \frac{2}{12} \]
\[ P(H) = \frac{1}{3} + \frac{5}{12} \]
To add these fractions, we find a common denominator (which is 12):
\[ P(H) = \frac{4}{12} + \frac{5}{12} = \frac{9}{12} = \frac{3}{4} \]

5. **Apply Bayes' Theorem**:
Now we can find \( P(C_1 | H) \):
\[ P(C_1 | H) = \frac{P(H | C_1) P(C_1)}{P(H)} \]
Substituting the values:
\[ P(C_1 | H) = \frac{1 \cdot \frac{1}{3}}{\frac{3}{4}} = \frac{\frac{1}{3}}{\frac{3}{4}} = \frac{1}{3} \cdot \frac{4}{3} = \frac{4}{9} \]

### Final Answer:
The probability that the coin chosen was the two-headed coin given that it shows heads is:
\[ \boxed{\frac{4}{9}} \]