Let D represent departing on time and A represent arriving on time.
P(D)=0.9
P(A)=0.8
P(D∩A)=0.75
a. P(A|D)
P(A∣D)=P(D)P(A∩D)=0.90.75=0.833
b. P(A∣Dc)
P(Dc)=1−0.9=0.1
P(A∩Dc)=0.1−(1−0.9−0.8+0.75)=0.05
P(A∣Dc)=P(Dc)P(A∩Dc)=0.10.05=0.5
c. The events departing on time and arriving on time are not independent. P(A∩D)=0.75=P(A)×P(D)=0.72
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