An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn, the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red is

Options

A. `(26)/(49)`
B. `(32)/(49)`
C. `(27)/(49)`
D. `(21)/(49)`

Answer

To solve the problem step by step, we will analyze the situation using probability concepts.

### Step 1: Identify the initial conditions
Initially, the urn contains:
- 5 red balls
- 2 green balls

Total number of balls = 5 + 2 = 7

### Step 2: Calculate the probabilities for the first draw
We will consider two cases based on the color of the ball drawn first.

**Case 1:** The first ball drawn is green.
**Case 2:** The first ball drawn is red.

#### Case 1: First ball is green
- Probability of drawing a green ball:
\[ P(\text{Green}) = \frac{2}{7} \]
- If a green ball is drawn, we add a red ball to the urn. The new composition will be:
- Red balls = 5 + 1 = 6
- Green balls = 2 - 1 = 1
- Total balls = 6 + 1 = 7

Now, we need to find the probability of drawing a red ball second:
- Probability of drawing a red ball:
\[ P(\text{Red | Green}) = \frac{6}{7} \]

#### Case 2: First ball is red
- Probability of drawing a red ball:
\[ P(\text{Red}) = \frac{5}{7} \]
- If a red ball is drawn, we add a green ball to the urn. The new composition will be:
- Red balls = 5 - 1 = 4
- Green balls = 2 + 1 = 3
- Total balls = 4 + 3 = 7

Now, we need to find the probability of drawing a red ball second:
- Probability of drawing a red ball:
\[ P(\text{Red | Red}) = \frac{4}{7} \]

### Step 3: Calculate the total probability of drawing a red ball second
Using the law of total probability, we can combine the probabilities from both cases:
\[ P(\text{Red second}) = P(\text{Green}) \cdot P(\text{Red | Green}) + P(\text{Red}) \cdot P(\text{Red | Red}) \]
Substituting the values we calculated:
\[ P(\text{Red second}) = \left(\frac{2}{7} \cdot \frac{6}{7}\right) + \left(\frac{5}{7} \cdot \frac{4}{7}\right) \]
Calculating each term:
\[ = \frac{12}{49} + \frac{20}{49} = \frac{32}{49} \]

### Final Answer
The probability that the second ball drawn is red is:
\[ \frac{32}{49} \]