Probability of three sequential events

Let $A_i$ be an event $i$-th police stop him. We are interested in $$P(A_1'\cap A_2'\cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 \cdot 0.5 \cdot 0.7 = 0.105$$

(since $A_1, A_2, A_3$ are independant so are $A_1', A_2', A_3'$ ) so your answer is correct.

You can think this in two ways. first way Not being caught on first check post and not being caught on second check post and not being caught on third post. i.e. $(1-P(A1))×(1-P(A2))×(1-P(A3) = 0.3×0.5×0.7 = 0.105$

second way Find P(being caught)

Caught on first check post Or Not being caught on first post and caught on second post Or Not being caught on first post and Not being caught on second post and caught on third post.

$P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3) =0.7 + 0.3×0.5 + 0.3×0.5×0.3 =0.895$

Now, P(not being caught) = 1- P(being caught) = 1- 0.895 =0.105