The possibilities are:

(1) Jo fails the theory part

(2) Jo passes the theory, but fails the practical

(3) Jo passes both

(A) The probability of (1) is .2

(B) The probability of (2) is (.8)(.4)=.32

(C) The probability of (3) is (.8)(.6)=.48

Note that the probabilites sum to 1 as expected.

**The probability that Jo only failed the practical part is .32**

** Let PT=pass theory and FP= failed practical. Then P(FP)=P(FP|PT)P(PT) or P(FP)=(.4)(.8)=.32

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