A small regional carrier accepted 14 reservations for a particular flight with 13 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a $41 \%$ chance, independently of each other. (Report answers accurate to 4 decimpl places.) Find the probability that overbooking occurs.

Step 1 ：We are given that a small regional carrier accepted 14 reservations for a particular flight with 13 seats. 11 reservations went to regular customers who will definitely arrive for the flight. The remaining 3 passengers each have a 41% chance of arriving, independently of each other.

Step 2 ：Overbooking occurs when more than 13 passengers arrive for the flight. Since 11 regular customers will definitely arrive, overbooking will occur if any of the remaining 3 passengers arrive.

Step 3 ：We need to calculate the probability that at least one of these 3 passengers arrives. Let's denote the number of these passengers as \(n = 3\) and the probability of each of them arriving as \(p = 0.41\).

Step 4 ：First, we calculate the probability that none of these 3 passengers arrive, which is \((1-p)^n = (1-0.41)^3 = 0.2054\).

Step 5 ：Then, the probability that at least one of these 3 passengers arrives, i.e., the probability that overbooking occurs, is \(1 - (1-p)^n = 1 - 0.2054 = 0.7946\).

Step 6 ：Final Answer: The probability that overbooking occurs is approximately \(\boxed{0.7946}\).