# Statistics exercise with bayes theorem with 4 events

I got confused with the probabilities:

ABC airport has $3$ terminals: $T_1$ is for international flights, $T_2$ just for national flights and $T_3$ for regional flights. From all flights departing from the airport $10\%$ are international and $30\%$ are national flights. The probability of an international flight having a delayed departure is $0.10$; from those flights having delayed departures $20\%$ are regional flights, the probability a flight has a delayed departure is $0.05$.

a) What is the probability that a national flight, randomly chosen, has a delayed departure? b) It is known that a specific flight has had a delayed departure. From which terminal is more probable it has departed?

I have already defined events ( but I'm not sure if they are correct) $P[T_1]= 0.1$; $P[T_2]= 0.3$; $P[T_3]= 0.6$; $P[D \mid T_1]= 0.1$; $P[T3\mid D]= 0.2$; $P[D]= 0.05$, while $D$ is an event that the flight was delayed. How to solve this excersise?

It is easy to see that $(D \cap T_i) \cap (D \cap T_j)= \emptyset$ for all $\mathcal i,j$ so these sets are a partition of D. Now by what we were given we compute:
• $P( D \cap T_1)= P( D| T_1) P(T_1)=0.01$
• $P(T_3 \cap D)= P(T_3|D) P(D)=0.01$
Now, we have (why?): $P(D)= P(D \cap T_1) + P(D \cap T_2) + P(D \cap T_3)$ So we get that $P(D \cap T_2) =0.03$. I hope this helped.