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 will use Bayes' theorem to find the probability that dice A is being used given that red has turned up in the first three throws.

### 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. **Calculate Prior Probabilities**:
- The probability of using dice A:
\[ P(A) = \frac{1}{2} \]
- The probability of using dice B:
\[ P(B) = \frac{1}{2} \]

3. **Calculate Conditional Probabilities**:
- For dice A, the probability of rolling red on one throw:
\[ P(R \mid A) = \frac{4}{6} = \frac{2}{3} \]
- For dice B, the probability of rolling red on one throw:
\[ P(R \mid B) = \frac{3}{6} = \frac{1}{2} \]

4. **Calculate Probability of Getting Red in Three Throws**:
- The probability of getting red in three consecutive throws with dice A:
\[ P(R \mid A)^{3} = \left(\frac{2}{3}\right)^{3} = \frac{8}{27} \]
- The probability of getting red in three consecutive throws with dice B:
\[ P(R \mid B)^{3} = \left(\frac{1}{2}\right)^{3} = \frac{1}{8} \]

5. **Calculate Total Probability of R**:
- Using the law of total probability:
\[ P(R) = P(R \mid A) P(A) + P(R \mid B) P(B) \]
- Substituting the values:
\[ P(R) = \left(\frac{8}{27} \cdot \frac{1}{2}\right) + \left(\frac{1}{8} \cdot \frac{1}{2}\right) \]
- Simplifying:
\[ P(R) = \frac{8}{54} + \frac{1}{16} \]
- Finding a common denominator (which is 432):
\[ P(R) = \frac{64}{432} + \frac{27}{432} = \frac{91}{432} \]

6. **Apply Bayes' Theorem**:
- We want to find \( P(A \mid R) \):
\[ P(A \mid R) = \frac{P(R \mid A) P(A)}{P(R)} \]
- Substituting the values:
\[ P(A \mid R) = \frac{\left(\frac{8}{27}\right) \cdot \left(\frac{1}{2}\right)}{\frac{91}{432}} \]
- Simplifying:
\[ P(A \mid R) = \frac{\frac{8}{54}}{\frac{91}{432}} = \frac{8 \cdot 432}{54 \cdot 91} = \frac{3456}{4902} \]
- Reducing the fraction:
\[ P(A \mid R) = \frac{64}{91} \]

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