Dice A has 4 red and 2 white faces whereas dice B has 3 red and 3 white faces. A coin is tossed once, if it falls head then the game continues by throwing the dice A and if it falls tail then the dice B is to be used. If red turns up at first 3 throws, then the probability that dice A is being used is

Options

A. `(7)/(37)`
B. `(64)/(91)`
C. `(9)/(41)`
D. `(27)/(35)`

Answer

To solve the problem, we need to find the probability that Dice A is being used given that red has turned up in the first three throws. We will use Bayes' Theorem for this calculation.

### Step-by-step Solution:

1. **Define Events**:
- Let \( A \) be the event that Dice A is used.
- Let \( B \) be the event that Dice B is used.
- Let \( R \) be the event that red turns up in the first three throws.

2. **Prior Probabilities**:
- The probability of selecting Dice A or Dice B after tossing the coin is:
\[ P(A) = P(B) = \frac{1}{2} \]

3. **Calculate Probabilities of Red for Each Die**:
- For Dice A (4 red and 2 white):
\[ P(R|A) = \left(\frac{4}{6}\right)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \]
- For Dice B (3 red and 3 white):
\[ P(R|B) = \left(\frac{3}{6}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

4. **Apply Bayes' Theorem**:
- According to Bayes' Theorem:
\[ P(A|R) = \frac{P(R|A) \cdot P(A)}{P(R|A) \cdot P(A) + P(R|B) \cdot P(B)} \]
- Substitute the values:
\[ P(A|R) = \frac{\frac{8}{27} \cdot \frac{1}{2}}{\frac{8}{27} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{2}} \]

5. **Simplify the Denominator**:
- Calculate \( P(R|A) \cdot P(A) + P(R|B) \cdot P(B) \):
\[ P(R|A) \cdot P(A) = \frac{8}{27} \cdot \frac{1}{2} = \frac{8}{54} \]
\[ P(R|B) \cdot P(B) = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16} \]
- Find a common denominator (which is 432):
\[ \frac{8}{54} = \frac{64}{432}, \quad \frac{1}{16} = \frac{27}{432} \]
- Thus, the denominator becomes:
\[ \frac{64}{432} + \frac{27}{432} = \frac{91}{432} \]

6. **Final Calculation**:
- Now substitute back into the formula:
\[ P(A|R) = \frac{\frac{8}{54}}{\frac{91}{432}} = \frac{8 \cdot 432}{54 \cdot 91} \]
- Simplifying gives:
\[ P(A|R) = \frac{3456}{4914} \]
- This simplifies to:
\[ P(A|R) = \frac{64}{91} \]

### Final Answer:
The probability that Dice A is being used given that red has turned up in the first three throws is:
\[ \boxed{\frac{64}{91}} \]