To solve the given problem, we need to find the probability that player A wins the game by rolling a sum of 5 before player B can roll a sum of 8. Let's break down the solution into clear steps:
1. **Possible Outcomes**: When a pair of dice is rolled, there are \(6 \times 6 = 36\) possible outcomes.
2. **Winning Conditions**: - A wins by rolling a sum of 5. - B wins by rolling a sum of 8.
3. **Calculating the Probability of Rolling a Specific Sum**:
4. **Probabilities**: - Probability that A rolls a sum of 5: \(\frac{4}{36} = \frac{1}{9}\) - Probability that B rolls a sum of 8: \(\frac{5}{36}\) - Probability that neither rolls a sum: \(1 - \frac{1}{9} - \frac{5}{36} = \frac{25}{36}\)
5. **Define the Events**: - Let \( p \) be the probability that A wins, starting with A's turn.
6. **Constructing the Equation**: - A could win immediately by rolling a sum of 5: \(p = \frac{1}{9} + \frac{25}{36}p\).
7. **Solve for \( p \)**:
8. **Conclusion**: The probability \( p_A \) that A wins, considering A throws first, is \(\frac{9}{19}\). Therefore, the correct answer is \(\frac{9}{19}\).
Given the following sums:
The possible pairs are \( (1,4), (2,3), (3,2) \), leading to:
\[ P(A) = \frac{4}{36}. \]
The possible pairs are \( (2,6), (3,5), (4,4), (5,3), (6,2) \), leading to:
\[ P(B) = \frac{5}{36}. \]
The probability of \( A \) winning is given by the series: \[ P(A \text{ wins}) = P(A) + P(A^C)P(B)P(A) + P(A^C)P(B)P(A^C)P(B) + \dots \] This series simplifies to: \[ P(A \text{ wins}) = \frac{9}{19}. \]
The probability of \( A \) winning is \( \frac{9}{19} \).
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]