Overbooking flight

A small regional carrier accepted 12 reservations for a particular flight with 11 seats. Seven reservations went to regular customers who would arrive for the flight. Each remaining passenger will arrive for the flight with a 49% chance, independently of each other.
(Report answers accurate to 4 decimal places. )

A. Find the probability that overbooking occurs.
B. Find the probability that the flight has empty seats.

Correct answer:

o =  0.0282
e =  0.8248

Step-by-step explanation:

p=49%=10049=0.49 r=7 n=12r=127=5 s=11r=117=4  s=4 seats  overbook: 5 arrive  OK: 4 arrive  free: 0,1,2,3 arrive  o=(5n)p5(1p)n5=(55)0.495(10.49)55=10.4950.510=0.0282
e1=(3n)p3(1p)n3=(35)0.493(10.49)53=100.4930.5120.306 e2=(2n)p2(1p)n2=(25)0.492(10.49)52=100.4920.5130.3185 e3=(1n)p1(1p)n1=(15)0.491(10.49)51=50.4910.5140.1657 e4=(0n)p0(1p)n0=(05)0.490(10.49)50=10.4900.5150.0345  e=e1+e2+e3+e4=0.306+0.3185+0.1657+0.0345=0.8248

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