A purse contains 4 copper and 3 silver coins. Another purse contains 6 copper and 2 silver coins. A coin is taken out from any purse, the probability that it is a silver coin, is

Options

A. `(37)/(56)`
B. `(19)/(56)`
C. `(4)/(7)`
D. `(2)/(3)`

Explanation

To solve the problem, we need to find the probability of drawing a silver coin from either of the two purses. We will use the law of total probability to calculate this.

### Step-by-Step Solution:

1. **Identify the Total Coins in Each Purse:**
- Purse 1 contains 4 copper coins and 3 silver coins. Thus, the total number of coins in Purse 1 is:
\[ 4 + 3 = 7 \text{ coins} \]
- Purse 2 contains 6 copper coins and 2 silver coins. Thus, the total number of coins in Purse 2 is:
\[ 6 + 2 = 8 \text{ coins} \]

2. **Calculate the Probability of Choosing Each Purse:**
- Since there are two purses, the probability of selecting either purse is equal:
\[ P(E1) = P(E2) = \frac{1}{2} \]

3. **Calculate the Probability of Drawing a Silver Coin from Each Purse:**
- For Purse 1:
- The probability of drawing a silver coin from Purse 1 is:
\[ P(A|E1) = \frac{\text{Number of Silver Coins in Purse 1}}{\text{Total Coins in Purse 1}} = \frac{3}{7} \]
- For Purse 2:
- The probability of drawing a silver coin from Purse 2 is:
\[ P(A|E2) = \frac{\text{Number of Silver Coins in Purse 2}}{\text{Total Coins in Purse 2}} = \frac{2}{8} = \frac{1}{4} \]

4. **Use the Law of Total Probability:**
- The total probability of drawing a silver coin (event A) can be calculated as:
\[ P(A) = P(E1) \cdot P(A|E1) + P(E2) \cdot P(A|E2) \]
- Substituting the values we calculated:
\[ P(A) = \left(\frac{1}{2} \cdot \frac{3}{7}\right) + \left(\frac{1}{2} \cdot \frac{1}{4}\right) \]

5. **Calculate Each Term:**
- For the first term:
\[ \frac{1}{2} \cdot \frac{3}{7} = \frac{3}{14} \]
- For the second term:
\[ \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \]

6. **Find a Common Denominator and Add:**
- The least common multiple of 14 and 8 is 56. We convert each fraction:
\[ \frac{3}{14} = \frac{3 \times 4}{14 \times 4} = \frac{12}{56} \]
\[ \frac{1}{8} = \frac{1 \times 7}{8 \times 7} = \frac{7}{56} \]
- Now, add the two fractions:
\[ P(A) = \frac{12}{56} + \frac{7}{56} = \frac{19}{56} \]

### Final Answer:
The probability that a coin drawn is a silver coin is:
\[ \boxed{\frac{19}{56}} \]