A critical orthopedic surgery is performed on 3 patients. The probability of recovering a patient is 0.6. Then the probability that after surgery, exactly two of them will recover is:
The probability of a patient recovering, \( P(E) = 0.6 \), and the probability of not recovering, \( P(\bar{E}) = 0.4 \). \[ P(E) = P(A) \cdot P(B) \cdot P(\bar{C}) + P(A) \cdot P(\bar{B}) \cdot P(C) + P(\bar{A}) \cdot P(B) \cdot P(C) \] Substituting the values for \( P(E) \) and \( P(\bar{E}) \): \[ P(E) = (0.6 \times 0.6 \times 0.4) + (0.6 \times 0.4 \times 0.6) + (0.4 \times 0.6 \times 0.6) \] \[ P(E) = 0.144 + 0.144 + 0.144 = 0.432 \] Thus, the total probability that at least one patient recovers is \({0.432}\).
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