A bag contains 16 coins of which two are counterfeit with heads on both sides. The rest are fair coins. One is selected at random from the bag and tossed. The probability of getting a head is

Options

A. `9//16`
B. `11//16`
C. `5//9`
D. none of these

Explanation

To solve the problem, we need to find the probability of getting a head when a coin is selected at random from a bag containing 16 coins, of which 2 are counterfeit (having heads on both sides) and the remaining 14 are fair coins.

### Step-by-Step Solution:

1. **Identify the total number of coins and their types**:
- Total coins = 16
- Counterfeit coins (2-headed) = 2
- Fair coins (1 head, 1 tail) = 16 - 2 = 14

2. **Calculate the probability of selecting a counterfeit coin**:
- Probability of selecting a counterfeit coin (Event A) = Number of counterfeit coins / Total number of coins
- \( P(A) = \frac{2}{16} = \frac{1}{8} \)

3. **Calculate the probability of getting a head given that a counterfeit coin is selected**:
- Since a counterfeit coin has heads on both sides, the probability of getting a head given that we have selected a counterfeit coin (Event B given A) = 1
- \( P(B|A) = 1 \)

4. **Calculate the probability of selecting a fair coin**:
- Probability of selecting a fair coin (Event A') = 1 - P(A)
- \( P(A') = 1 - \frac{1}{8} = \frac{7}{8} \)

5. **Calculate the probability of getting a head given that a fair coin is selected**:
- A fair coin has one head and one tail, so the probability of getting a head given that we have selected a fair coin (Event B given A') = 0.5
- \( P(B|A') = \frac{1}{2} \)

6. **Use the law of total probability to find the overall probability of getting a head**:
- The total probability of getting a head (Event B) can be calculated using:
\[ P(B) = P(A) \cdot P(B|A) + P(A') \cdot P(B|A') \]
- Substituting the values we calculated:
\[ P(B) = \left(\frac{1}{8} \cdot 1\right) + \left(\frac{7}{8} \cdot \frac{1}{2}\right) \]
- Simplifying this:
\[ P(B) = \frac{1}{8} + \frac{7}{16} \]
- To add these fractions, convert \(\frac{1}{8}\) to have a common denominator of 16:
\[ P(B) = \frac{2}{16} + \frac{7}{16} = \frac{9}{16} \]

### Final Answer:
The probability of getting a head when a coin is selected at random from the bag is \( \frac{9}{16} \).