A box contain three coins, one coin is fair, one coin is two-headed, and one coin is weighted so that the probability of heads, appearing is 1/3 . A coin is selected at random and tossed. Then the probability that heads appears is

Options

A. `(11)/(18)`
B. `(12)/(18)`
C. `(13)/(18)`
D. `(14)/(18)`

Answer

To solve the problem step by step, we will calculate the probability of getting heads when a coin is selected at random from a box containing three different coins: a fair coin, a two-headed coin, and a weighted coin.

### Step 1: Define the coins and their probabilities of heads
- Let \( F \) be the fair coin: Probability of heads \( P(H|F) = \frac{1}{2} \)
- Let \( T \) be the two-headed coin: Probability of heads \( P(H|T) = 1 \)
- Let \( W \) be the weighted coin: Probability of heads \( P(H|W) = \frac{1}{3} \)

### Step 2: Determine the probability of selecting each coin
Since there are three coins and they are selected at random, the probability of selecting each coin is:
- \( P(F) = \frac{1}{3} \)
- \( P(T) = \frac{1}{3} \)
- \( P(W) = \frac{1}{3} \)

### Step 3: Use the law of total probability to find the overall probability of heads
The total probability of getting heads \( P(H) \) can be calculated using the formula:
\[ P(H) = P(F) \cdot P(H|F) + P(T) \cdot P(H|T) + P(W) \cdot P(H|W) \]

Substituting the values we have:
\[ P(H) = \left(\frac{1}{3} \cdot \frac{1}{2}\right) + \left(\frac{1}{3} \cdot 1\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right) \]

### Step 4: Calculate each term
1. For the fair coin:
\[ P(F) \cdot P(H|F) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \]
2. For the two-headed coin:
\[ P(T) \cdot P(H|T) = \frac{1}{3} \cdot 1 = \frac{1}{3} \]
3. For the weighted coin:
\[ P(W) \cdot P(H|W) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} \]

### Step 5: Combine the probabilities
Now, we need to add these probabilities together:
\[ P(H) = \frac{1}{6} + \frac{1}{3} + \frac{1}{9} \]

### Step 6: Find a common denominator
The least common multiple of 6, 3, and 9 is 18. We will convert each fraction:
- \( \frac{1}{6} = \frac{3}{18} \)
- \( \frac{1}{3} = \frac{6}{18} \)
- \( \frac{1}{9} = \frac{2}{18} \)

### Step 7: Add the fractions
Now we can add them:
\[ P(H) = \frac{3}{18} + \frac{6}{18} + \frac{2}{18} = \frac{3 + 6 + 2}{18} = \frac{11}{18} \]

### Final Answer
The probability that heads appears when a coin is selected at random and tossed is:
\[ \boxed{\frac{11}{18}} \]