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**HallsofIvy** Another way of looking at it: writing "P" or "F" for pass or fail, and "Math", "French" and "History" in that order, there are $\displaystyle 2^3$ possible outcomes, with probabilities:

PPP: (0.7)(0.8)(0.6)= 0.336

PPF: (0.7)(0.8)(0.4)= 0.224

PFP: (0.7)(0.2)(0.6)= 0.084

PFF: (0.7)(0.2)(0.4)= 0.056

FPP: (0.3)(0.8)(0.6)= 0.144

FPF: (0.3)(0.8)(0.4)= 0.096

FFP: (0.3)(0.2)(0.6)= 0.036

FFF: (0.3)(0.2)(0.4)= 0.024

Of those, PPF, PFP, and FPP correspond to "fails just one exam" and they have total probability 0.224+ 0.084+ 0.144= 0.452.

That one exam being French is PFP which has probability 0.084 so the probability that she will fail French given that she fails just one exam is $\displaystyle \dfrac{0.084}{0.452}= 0.1858$