Answer:
(a) The probability of overbooking is 0.2135.
(b) The probability that the flight has empty seats is 0.4625.
Step-by-step explanation:
Let the random variable <em>X</em> represent the number of passengers showing up for the flight.
It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.
Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.
Number of reservations = 19
Regular customers = 14
Seats available = 17 - 14 = 3
Remaining reservations, n = 19 - 14 = 5
P (A remaining passenger will arrive), <em>p</em> = 0.52
The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.
(1)
Compute the probability of overbooking as follows:
P (Overbooking occurs) = P(More than 3 shows up for the flight)
![=P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135](https://tex.z-dn.net/?f=%3DP%28X%3E3%29%5C%5C%5C%5C%3D%7B5%5Cchoose%204%7D%280.52%29%5E%7B4%7D%281-0.52%29%5E%7B5-4%7D%2B%7B5%5Cchoose%205%7D%280.52%29%5E%7B5%7D%281-0.52%29%5E%7B5-5%7D%5C%5C%5C%5C%3D0.175478784%2B0.0380204032%5C%5C%5C%5C%3D0.2134991872%5C%5C%5C%5C%5Capprox%200.2135)
Thus, the probability of overbooking is 0.2135.
(2)
Compute the probability that the flight has empty seats as follows:
P (The flight has empty seats) = P (Less than 3 shows up for the flight)
![=P(X](https://tex.z-dn.net/?f=%3DP%28X%3C3%29%5C%5C%5C%5C1-P%28X%5Cgeq%203%29%5C%5C%5C%5C%3D1-%5B%7B5%5Cchoose%203%7D%280.52%29%5E%7B3%7D%281-0.52%29%5E%7B5-3%7D%2B%7B5%5Cchoose%204%7D%280.52%29%5E%7B4%7D%281-0.52%29%5E%7B5-4%7D%2B%7B5%5Cchoose%205%7D%280.52%29%5E%7B5%7D%281-0.52%29%5E%7B5-5%7D%5D%5C%5C%5C%5C%3D1-%5B0.323960832%2B0.175478784%2B0.0380204032%5D%5C%5C%5C%5C%3D0.4625399808%5C%5C%5C%5C%5Capprox%200.4625)
Thus, the probability that the flight has empty seats is 0.4625.