A bag contains 3 red and 4 black balls and another bag has 4 red and 2 black balls . One bag is selected at random and from the selected at random and from the selected bag a ball is drawn .Let E be the event the first bag is selected F be the event that the second bag is selected ,G be the event that ball drawn is red .
find `P(G/E )`

Options

A. `5/6`
B. `5/14`
C. `3/7`
D. none of these

Answer

To solve the problem step by step, let's define the events and calculate the required probability.

### Step 1: Define the Events
- Let \( E \) be the event that the first bag is selected.
- Let \( F \) be the event that the second bag is selected.
- Let \( G \) be the event that a red ball is drawn.

### Step 2: Calculate the Probability of Selecting Each Bag
There are 2 bags, and one is selected at random. Therefore:
- \( P(E) = \frac{1}{2} \) (probability of selecting the first bag)
- \( P(F) = \frac{1}{2} \) (probability of selecting the second bag)

### Step 3: Calculate the Probability of Drawing a Red Ball from the First Bag
The first bag contains 3 red balls and 4 black balls, making a total of 7 balls.
- The probability of drawing a red ball from the first bag is:
\[ P(G | E) = \frac{\text{Number of red balls in the first bag}}{\text{Total number of balls in the first bag}} = \frac{3}{7} \]

### Step 4: Find the Conditional Probability \( P(G | E) \)
We want to find \( P(G | E) \), which is already calculated in Step 3. Thus:
\[ P(G | E) = \frac{3}{7} \]

### Final Answer
The probability of drawing a red ball given that the first bag is selected is:
\[ \boxed{\frac{3}{7}} \]