Here given,
probability that a flight arrives on time = P(A) = 0.7
probability that a flight departs on time = P(B) = 0.8.
probability that a flight neither arrives on time nor departs on time = P(A' "\\bigcap" B') = 0.15
probability that flight arrives on time and departs on time =P(A "\\bigcap" B)
Using De'morgan law P(A "\\bigcup" B)' = P(A' "\\bigcap" B')
so, P(A "\\bigcup" B)' = P(A' "\\bigcap" B') = 0.15
P(A "\\bigcup" B)' = 1- P(A "\\bigcup" B)
P(A "\\bigcup" B) = 1-P(A "\\bigcup" B)'
=1-0.15
=0.85
Now,
P(A "\\bigcup" B) = P(A) + P(B) -P(A "\\bigcap" B)
P(A "\\bigcap" B) = P(A) + P(B) - P(A "\\bigcup" B)
= 0.7 + 0.8 - 0.85
P(A "\\bigcap" B) = 0.65
ANSWER : probability that flight arrives on time and departs on time =P(A "\\bigcap" B)=0.65
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