Following are the questions based on two statements and answer the following based on the given statements.
A bag contains total 12 balls in which there are 5 green balls and rest are blue and red balls. What is difference between blue &red balls.
Statement I. If one ball taken out from bag probability of being either red or blue is `(7)/(12).`
Statement II. If two balls taken out from bag probability of being either red or blue is `(1)/(6).`

Options

A. Only statement II is sufficient
B. Either statement I or Statement Il alone is sufficient
C. Statement I and Il both together is sufficient
D. Only statement I is sufficient

Answer

To solve the problem, we need to determine the difference between the number of blue and red balls in a bag that contains a total of 12 balls, of which 5 are green. The remaining balls are either blue or red. We will analyze the two statements provided to see if they give us sufficient information to answer the question.

### Step 1: Determine the total number of blue and red balls
Given:
- Total balls = 12
- Green balls = 5

Thus, the total number of blue and red balls combined is:
\[ \text{Blue balls} + \text{Red balls} = 12 - 5 = 7 \]

Let:
- Number of blue balls = \( x \)
- Number of red balls = \( 7 - x \)

### Step 2: Analyze Statement I
**Statement I:** The probability of drawing either a red or blue ball when one ball is taken out is \( \frac{7}{12} \).

The probability of drawing a blue ball is:
\[ P(\text{Blue}) = \frac{x}{12} \]

The probability of drawing a red ball is:
\[ P(\text{Red}) = \frac{7 - x}{12} \]

Thus, the combined probability of drawing either a red or blue ball is:
\[ P(\text{Red or Blue}) = P(\text{Blue}) + P(\text{Red}) = \frac{x + (7 - x)}{12} = \frac{7}{12} \]

Since this statement is true, it does not provide any new information about the specific values of \( x \) and \( 7 - x \). Therefore, we cannot determine the difference between blue and red balls from this statement alone.

### Step 3: Analyze Statement II
**Statement II:** The probability of drawing either a red or blue ball when two balls are taken out is \( \frac{1}{6} \).

The probability of drawing two balls (either red or blue) can be calculated using combinations:
1. Probability of drawing two blue balls:
\[ P(\text{2 Blue}) = \frac{x}{12} \cdot \frac{x - 1}{11} \]

2. Probability of drawing two red balls:
\[ P(\text{2 Red}) = \frac{7 - x}{12} \cdot \frac{6 - x}{11} \]

3. Probability of drawing one blue and one red ball (in either order):
\[ P(\text{1 Blue, 1 Red}) = \frac{x}{12} \cdot \frac{7 - x}{11} + \frac{7 - x}{12} \cdot \frac{x}{11} = 2 \cdot \frac{x(7 - x)}{12 \cdot 11} \]

Thus, the total probability of drawing either red or blue when two balls are taken out is:
\[ P(\text{Red or Blue}) = P(\text{2 Blue}) + P(\text{2 Red}) + P(\text{1 Blue, 1 Red}) \]

Setting this equal to \( \frac{1}{6} \):
\[ \frac{x(x - 1)}{12 \cdot 11} + \frac{(7 - x)(6 - x)}{12 \cdot 11} + 2 \cdot \frac{x(7 - x)}{12 \cdot 11} = \frac{1}{6} \]

### Step 4: Solve the equation
Multiplying through by \( 12 \cdot 11 \) to eliminate the denominators:
\[ x(x - 1) + (7 - x)(6 - x) + 2x(7 - x) = 22 \]

Expanding and simplifying:
1. \( x^2 - x + (42 - 13x + x^2) + (14x - 2x^2) = 22 \)
2. Combine like terms:
\[ 2x^2 - 14x + 42 = 22 \]
3. Rearranging gives:
\[ 2x^2 - 14x + 20 = 0 \]
4. Dividing by 2:
\[ x^2 - 7x + 10 = 0 \]
5. Factoring:
\[ (x - 5)(x - 2) = 0 \]
Thus, \( x = 5 \) or \( x = 2 \).

### Step 5: Calculate the difference
1. If \( x = 5 \) (blue balls), then red balls = \( 7 - 5 = 2 \). Difference = \( 5 - 2 = 3 \).
2. If \( x = 2 \) (blue balls), then red balls = \( 7 - 2 = 5 \). Difference = \( 5 - 2 = 3 \).

### Conclusion
From Statement II, we can conclude that the difference between the number of blue and red balls is 3. Therefore, the answer is that only Statement II is sufficient to answer the question.