Study the given information and answer the following questions.
There are three bags A, B and C. In each bag there are three types of coloured balls - Yellow, Green and Black In bag A the number of Yellow balls is y and the number of Green balls is g. The number of green balls is 4 more than the number of Yellow balls. When one ball is picked at random then the probability of getting Black ball is `5//13` . The value of y is `18 2/11%` less than g. In bag B, the number of Yellow balls is `22 2/9%` more than that of bag A. If two balls are picked at random from bag B then the probability of getting both Green colour balls is `4//37`. The total number of balls in bag B is 75. In bag C, the ratio of the number of Green balls to the number of Black balls is 7:5. The total number of Green and Black balls is 36. If one of the balls, is picked at random t hen the probability of getting one Yellow ball is `7//13`.
If one ball is picked at random from each of the bags A and B then find the probability that both the balls are of same colour

Options

A. `((21 X 47))/((65 X 75))`
B. `((22 X 43))/((65 X 75))`
C. ` ((11 X 17))/((65 X 75))`
D. None of these

Answer

To solve the problem step by step, we will analyze the information given for each bag (A, B, and C) and calculate the required probabilities.

### Step 1: Analyze Bag A
1. Let the number of Yellow balls in Bag A be \( y \) and the number of Green balls be \( g \).
2. According to the problem, \( g = y + 4 \).
3. The probability of picking a Black ball from Bag A is given as \( \frac{5}{13} \).
4. The total number of balls in Bag A is \( y + g + b \) (where \( b \) is the number of Black balls).
5. From the probability, we have:
\[ P(\text{Black}) = \frac{b}{y + g + b} = \frac{5}{13} \]
This implies:
\[ b = \frac{5}{13} (y + g + b) \]
Rearranging gives:
\[ 13b = 5(y + g + b) \implies 13b = 5y + 5g + 5b \implies 8b = 5y + 5g \]
Thus:
\[ b = \frac{5}{8}(y + g) \]

6. We also know \( y \) is \( 18 \frac{2}{11}\% \) less than \( g \). This can be expressed as:
\[ y = g - \frac{200}{11} \cdot \frac{g}{100} = g - \frac{2g}{11} = \frac{9g}{11} \]
Therefore, we can substitute \( g \) in terms of \( y \):
\[ g = \frac{11}{9}y \]

7. Substituting \( g = y + 4 \) into \( g = \frac{11}{9}y \):
\[ y + 4 = \frac{11}{9}y \implies 4 = \frac{11}{9}y - y = \frac{2}{9}y \implies y = 18 \]
Thus, \( g = 22 \) and \( b = \frac{5}{8}(y + g) = \frac{5}{8}(18 + 22) = \frac{5}{8} \cdot 40 = 25 \).

### Step 2: Analyze Bag B
1. The number of Yellow balls in Bag B is \( 22 \frac{2}{9}\% \) more than in Bag A:
\[ \text{Increase} = \frac{200}{9} \cdot \frac{18}{100} = \frac{36}{9} = 4 \]
Therefore, the number of Yellow balls in Bag B is:
\[ 18 + 4 = 22 \]

2. The total number of balls in Bag B is 75, and the probability of picking two Green balls is \( \frac{4}{37} \):
\[ P(\text{Green}) = \frac{g_B}{75} \cdot \frac{g_B - 1}{74} = \frac{4}{37} \]
Let \( g_B \) be the number of Green balls in Bag B. Thus:
\[ g_B(g_B - 1) = \frac{4}{37} \cdot 75 \cdot 74 = 4 \cdot 150 = 600 \]
This leads to the quadratic equation:
\[ g_B^2 - g_B - 600 = 0 \]
Solving gives \( g_B = 25 \) (the positive root).

3. The number of Black balls in Bag B is:
\[ 75 - (22 + 25) = 28 \]

### Step 3: Analyze Bag C
1. The ratio of Green to Black balls is \( 7:5 \) and the total is 36:
\[ 7a + 5a = 36 \implies 12a = 36 \implies a = 3 \]
Thus, Green balls = \( 21 \) and Black balls = \( 15 \).

2. The total number of balls in Bag C is \( 21 + 15 + y_C \) (where \( y_C \) is the number of Yellow balls). The probability of picking a Yellow ball is \( \frac{7}{13} \):
\[ P(\text{Yellow}) = \frac{y_C}{21 + 15 + y_C} = \frac{7}{13} \]
Solving gives:
\[ 13y_C = 7(36 + y_C) \implies 13y_C = 252 + 7y_C \implies 6y_C = 252 \implies y_C = 42 \]

### Step 4: Find the Probability of Picking Same Color Balls
1. The probabilities of picking the same color balls from Bags A and B:
- Yellow: \( \frac{18}{65} \cdot \frac{22}{75} \)
- Green: \( \frac{22}{65} \cdot \frac{25}{75} \)
- Black: \( \frac{25}{65} \cdot \frac{28}{75} \)

2. Total probability:
\[ P(\text{Same Color}) = \left( \frac{18}{65} \cdot \frac{22}{75} \right) + \left( \frac{22}{65} \cdot \frac{25}{75} \right) + \left( \frac{25}{65} \cdot \frac{28}{75} \right) \]

3. Calculating each term:
- Yellow: \( \frac{396}{4875} \)
- Green: \( \frac{550}{4875} \)
- Black: \( \frac{700}{4875} \)

4. Adding these:
\[ P(\text{Same Color}) = \frac{396 + 550 + 700}{4875} = \frac{1646}{4875} \]

### Final Answer
The probability that both balls are of the same color when one ball is picked from each of the bags A and B is \( \frac{1646}{4875} \).