Question 1200769: When admitted to the Chess club, Tomas has to play three games with club members A, B, and win at least two games in a row. The probabilities of Tomas losing against A and B are pA and pB, respectively. The outcomes of the games are independent of each other. Tomas can choose only one opponent order: ABA or BAB. What is the probability of winning in the club in the first and second cases? ([pA, pB] = [0.47, 0.53])
Answer by asinus(45)   (Show Source):
You can put this solution on YOUR website! **1. Define Probabilities:**
* **wA:** Probability of winning against A = 1 - pA = 1 - 0.47 = 0.53
* **wB:** Probability of winning against B = 1 - pB = 1 - 0.53 = 0.47
**2. Calculate Probabilities for Each Order:**
* **ABA Order:**
* **Winning Scenario:** Win A, Win B, Win/Lose A
* Probability: wA * wB * (wA + pA) = 0.53 * 0.47 * (0.53 + 0.47) = 0.2491
* **BAB Order:**
* **Winning Scenario:** Win B, Win A, Win/Lose B
* Probability: wB * wA * (wB + pB) = 0.47 * 0.53 * (0.47 + 0.53) = 0.2491
**3. Conclusion:**
* The probability of winning with the ABA order is 0.2491.
* The probability of winning with the BAB order is 0.2491.
**Therefore, both orders have the same probability of winning.**
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