Respuesta :
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 X 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), p = 0.52
The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.
(1)
Compute the probability of overbooking as follows:
P (Overbooking occurs) = P(More than 3 shows up for the flight)
[tex]=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[/tex]
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)
[tex]=P(X<3)\\\\1-P(X\geq 3)\\\\=1-[{5\choose 3}(0.52)^{3}(1-0.52)^{5-3}+{5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}]\\\\=1-[0.323960832+0.175478784+0.0380204032]\\\\=0.4625399808\\\\\approx 0.4625[/tex]
Thus, the probability that the flight has empty seats is 0.4625.
