A bag contains three coins, one of which has head on both sides, another is a biased coin that shows up heads 90% of the time and the third one is an unbiased coin. A coin is taken out from the bag at random and tossed. If it is shows up a head, then the probability that it is the unbiased coin, is:

Options

A. `3//8`
B. `5//12`
C. `5//25`
D. `1//3`

Answer

To solve the problem step by step, we will use Bayes' theorem. The goal is to find the probability that the coin is unbiased given that it shows heads.

### Step 1: Define the Events
Let:
- \( U \): Event that the coin is unbiased.
- \( B \): Event that the coin is biased (90% heads).
- \( C \): Event that the coin has heads on both sides.
- \( H \): Event that the coin shows heads.

We need to find \( P(U | H) \).

### Step 2: Apply Bayes' Theorem
Using Bayes' theorem:
\[ P(U | H) = \frac{P(H | U) \cdot P(U)}{P(H)} \]

### Step 3: Calculate \( P(H | U) \)
For the unbiased coin:
\[ P(H | U) = \frac{1}{2} \quad \text{(since it has equal chances of heads and tails)} \]

### Step 4: Calculate \( P(U) \)
Since there are three coins, the probability of selecting the unbiased coin is:
\[ P(U) = \frac{1}{3} \]

### Step 5: Calculate \( P(H) \)
We need to calculate the total probability of getting heads from all three coins:
\[ P(H) = P(H | U) \cdot P(U) + P(H | B) \cdot P(B) + P(H | C) \cdot P(C) \]

- For the biased coin:
\[ P(H | B) = \frac{9}{10} \]
- For the coin with heads on both sides:
\[ P(H | C) = 1 \]

Now, we can substitute these values:
\[ P(H) = \left(\frac{1}{2} \cdot \frac{1}{3}\right) + \left(\frac{9}{10} \cdot \frac{1}{3}\right) + \left(1 \cdot \frac{1}{3}\right) \]

Calculating each term:
\[ P(H) = \frac{1}{6} + \frac{9}{30} + \frac{1}{3} \]

Converting \(\frac{1}{3}\) to a common denominator of 30:
\[ \frac{1}{3} = \frac{10}{30} \]

Now, substituting back:
\[ P(H) = \frac{1}{6} + \frac{9}{30} + \frac{10}{30} = \frac{5}{30} + \frac{9}{30} + \frac{10}{30} = \frac{24}{30} = \frac{4}{5} \]

### Step 6: Substitute Back into Bayes' Theorem
Now we can substitute back into Bayes' theorem:
\[ P(U | H) = \frac{P(H | U) \cdot P(U)}{P(H)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{5}} \]

Calculating the numerator:
\[ \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \]

Now substituting:
\[ P(U | H) = \frac{\frac{1}{6}}{\frac{4}{5}} = \frac{1}{6} \cdot \frac{5}{4} = \frac{5}{24} \]

### Final Answer
Thus, the probability that the coin is unbiased given that it shows heads is:
\[ \boxed{\frac{5}{24}} \]