Question

A small regional carrier accepted 17 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 57% chance, independently of each other. (Report answers accurate to 4 decimal places.) Find the probability that overbooking occurs. Find the probability that the flight has empty seats. Question Help: D Video D Post to forum Submit Question

Answer 1

Sos from the given information, Number of reservations accepted for Number of seats a flight = 17 in the flight = 13 available Number of reservations went to regular customers who \\ :بی arrive for the flight = 11 And each of the goreon remaining passengers will for the flight with a 57% Chance independently arrive each other. So here a have 2 seats remaining and reservations remain dining are 17-11=6 let X be random variable denotes the number of pamengers arrive out of remaining reservations. p=6.57

XN Binomial (n=6, p=0.57) The 6-2 ; x=0, 1, 2.-6 a) probability mass function of x is, P(X=x)() (0.57° C.-0.57) For overbooking to occur, the number of pomengers arrive for flight should be greater than So, we need to find here P( X2) 2 out of 6 P(x>2) = 1- P(x 52) = 1- [PCX=o) + P(X=1} + P(X=2>] [15)(0.52% (0.433 (0.435'+ (5) (0.54)(0435 -1- (0.00632136305+0.05 027688751+0.16661526674] + () (0.54;(0.49 = 1 0.22321351724 o 77678648271 0.7768 seat out b) for flight to have empty seats, if less than of 6 arrive. So, we need to find here P(X<2) P(X=o) + P(x=1) P(X22) = 1 5 6 (3) (0.57 (0.43) + (6) 10.57) (0.43) 0.00632136305+ 0.050 2768875 -0.05659825055 0.0566