Solution
= Solution
Starting from the airport, after one step the distribution is $(0,1/2,1/2)$. Multiplying once more by $P$ gives
$$
\boxed{\mathbb P(X_2=A)=\frac34,
\qquad
\mathbb P(X_2=B)=\mathbb P(X_2=C)=\frac18}.
$$
= Solution
Starting from the airport, after one step the distribution is $(0,1/2,1/2)$. Multiplying once more by $P$ gives
$$
\boxed{\mathbb P(X_2=A)=\frac34,
\qquad
\mathbb P(X_2=B)=\mathbb P(X_2=C)=\frac18}.
$$