Question:

**Step 1:** Identify qualifying cases. A student qualifies if they pass at least two out of three exams. The possible sequences of outcomes that lead to qualification are: * Pass, Pass, Pass (PPP) * Pass, Pass, Fail (PPF) * Pass, Fail, Pass (PFP) * Fail, Pass, Pass (FPP) **Step 2:** Calculate the probability for Case 1 (PPP). The probability of passing the first exam is $p$. Since no failure has occurred, the probability of passing the second exam remains $p$. Similarly, since no failure has occurred, the probability of passing the third exam remains $p$. $$P(PPP) = p \cdot p \cdot p = p^3$$ **Step 3:** Calculate the probability for Case 2 (PPF). The probability of passing the first exam is $p$. Since no failure has occurred, the probability of passing the second exam remains $p$. Since no failure has occurred, the probability of failing the third exam is $(1-p)$. $$P(PPF) = p \cdot p \cdot (1-p) = p^2(1-p)$$ **Step 4:** Calculate the probability for Case 3 (PFP). The probability of passing the first exam is $p$. Since no failure has occurred, the probability of failing the second exam is $(1-p)$. A failure occurred in the second exam, so the probability of passing the third exam becomes $p/2$. $$P(PFP) = p \cdot (1-p) \cdot \left(\frac{p}{2}\right) = \frac{p^2(1-p)}{2}$$ **Step 5:** Calculate the probability for Case 4 (FPP). The probability of failing the first exam is $(1-p)$. A failure occurred in the first exam, so the probability of passing the second exam becomes $p/2$. Since the second exam was passed (not failed), the condition "otherwise it remains the same" applies, meaning the probability of passing the third exam reverts to $p$. $$P(FPP) = (1-p) \cdot \left(\frac{p}{2}\right) \cdot p = \frac{p^2(1-p)}{2}$$ **Step 6:** Calculate the total probability of qualifying. The total probability of qualifying is the sum of the probabilities of these four mutually exclusive cases. $$P_{total} = P(PPP) + P(PPF) + P(PFP) + P(FPP)$$ $$P_{total} = p^3 + p^2(1-p) + \frac{p^2(1-p)}{2} + \frac{p^2(1-p)}{2}$$ $$P_{total} = p^3 + p^2(1-p) + p^2(1-p)$$ $$P_{total} = p^3 + 2p^2(1-p)$$ $$P_{total} = p^3 + 2p^2 - 2p^3$$ $$P_{total} = 2p^2 - p^3$$