Question

A small regional carrier accepted 20 reservations for a particular flight with 17 seats. 13 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 40% chance, independently of each other.
Find the probability that overbooking occurs.
Find the probability that the flight has empty seats.

Assume that a procedure yields a binomial distribution with a trial repeated n=5n=5 times. Use some form of technology to find the probability distribution given the probability p=0.299p=0.299 of success on a single trial.
(Report answers accurate to 4 decimal places.)

k	P(X = k)
0
1
2
3
4
5

Answer 1

1)

total seats=17

seats reserved =13

remaining passengers =20-13 = 7

probability that overbooking occurs =P(out of 7 passenger, more than 4 passenger arrive) =P(X=5) + P(X=6) + P(X=7) =

P ( X = 5) = C (7,5) * 0.4^5 * ( 1 - 0.4)^2=      0.0774
P ( X = 6) = C (7,6) * 0.4^6 * ( 1 - 0.4)^1=      0.0172
P ( X = 7) = C (7,7) * 0.4^7 * ( 1 - 0.4)^0=      0.0016

so, P(overbooking) = 0.0963
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P( flight has empty seats) = P(out of 7 passengers, less than 4 passenger arrive)

P ( X = 0) = C (7,0) * 0.4^0 * ( 1 - 0.4)^7=      0.0280
P ( X = 1) = C (7,1) * 0.4^1 * ( 1 - 0.4)^6=      0.1306
P ( X = 2) = C (7,2) * 0.4^2 * ( 1 - 0.4)^5=      0.2613
P ( X = 3) = C (7,3) * 0.4^3 * ( 1 - 0.4)^4=   0.2903

P( flight has empty seats) =P(X=0) + P(X=1) + P(X=2) + P(X3) =0.7102

==============================

Sample size , n =    5
Probability of an event of interest, p =   0.299

and probability is given by

P(X=x) = C(n,x)*px*(1-p)(n-x)

	X	P(X)
P ( X = 0) = C (5,0) * 0.299^0 * ( 1 - 0.299)^5=	0	0.1693
P ( X = 1) = C (5,1) * 0.299^1 * ( 1 - 0.299)^4=	1	0.3610
P ( X = 2) = C (5,2) * 0.299^2 * ( 1 - 0.299)^3=	2	0.3080
P ( X = 3) = C (5,3) * 0.299^3 * ( 1 - 0.299)^2=	3	0.1314
P ( X = 4) = C (5,4) * 0.299^4 * ( 1 - 0.299)^1=	4	0.0280
P ( X = 5) = C (5,5) * 0.299^5 * ( 1 - 0.299)^0=	5	0.0024