Raj and Sanchita are playing game in which they throw two dice alternately till one of them gets 9. Which one of the following could be the probability that Sanchita win the game?

Options

A. `7/15 or 8/15`
B. `6/11 or 5/11`
C. `8/17 or 9/17`
D. None of these

Answer

To solve the problem of determining the probability that Sanchita wins the game when she and Raj are throwing two dice alternately until one of them gets a sum of 9, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Calculate the Probability of Getting a Sum of 9:**
- When throwing two dice, the possible outcomes for each die are from 1 to 6.
- The combinations that yield a sum of 9 are:
- (3, 6)
- (4, 5)
- (5, 4)
- (6, 3)
- Thus, there are 4 favorable outcomes.
- The total number of outcomes when throwing two dice is \(6 \times 6 = 36\).
- Therefore, the probability \(P\) of getting a sum of 9 is:
\[ P(9) = \frac{4}{36} = \frac{1}{9} \]

2. **Probability of Not Getting a Sum of 9:**
- The probability of not getting a sum of 9 is:
\[ P(\text{not } 9) = 1 - P(9) = 1 - \frac{1}{9} = \frac{8}{9} \]

3. **Case 1: Sanchita Starts First:**
- If Sanchita rolls first, she can win immediately with a probability of \(\frac{1}{9}\).
- If she does not win, Raj rolls next. If he also does not win (with probability \(\frac{8}{9}\)), it returns to Sanchita.
- This process continues, forming a geometric series.
- The probability that Sanchita wins can be expressed as:
\[ P(Sanchita \text{ wins}) = \frac{1}{9} + \left(\frac{8}{9}\right) \left(\frac{8}{9}\right) P(Sanchita \text{ wins}) \]
- Let \(x\) be the probability that Sanchita wins. We can write:
\[ x = \frac{1}{9} + \left(\frac{8}{9}\right)^2 x \]
- Rearranging gives:
\[ x - \left(\frac{8}{9}\right)^2 x = \frac{1}{9} \]
\[ x \left(1 - \frac{64}{81}\right) = \frac{1}{9} \]
\[ x \cdot \frac{17}{81} = \frac{1}{9} \]
\[ x = \frac{1}{9} \cdot \frac{81}{17} = \frac{9}{17} \]

4. **Case 2: Raj Starts First:**
- If Raj rolls first, the probability that he wins immediately is \(\frac{1}{9}\).
- If he does not win, Sanchita rolls next. The probability that Sanchita wins can be expressed similarly:
\[ P(Sanchita \text{ wins}) = \left(\frac{8}{9}\right) \cdot \frac{1}{9} + \left(\frac{8}{9}\right) \cdot \left(\frac{8}{9}\right) P(Sanchita \text{ wins}) \]
- Let \(y\) be the probability that Sanchita wins in this case:
\[ y = \left(\frac{8}{9}\right) \cdot \frac{1}{9} + \left(\frac{8}{9}\right)^2 y \]
- Rearranging gives:
\[ y - \left(\frac{8}{9}\right)^2 y = \frac{8}{81} \]
\[ y \cdot \frac{17}{81} = \frac{8}{81} \]
\[ y = \frac{8}{17} \]

5. **Final Result:**
- The probabilities that Sanchita wins, depending on who starts, are \(\frac{9}{17}\) and \(\frac{8}{17}\).
- Therefore, the possible probabilities that Sanchita could win the game are \(\frac{9}{17}\) or \(\frac{8}{17}\).