Question

1. An airport limousine can accommodate up to 4 passengers on any one trip. The company will accept a maximum of 6 reservations for a trip, and a passenger must have a reservation. From previous records, 20% of all those making reservations do not show up for the trip. Answer the following questions assuming independence wherever appropriate.

   A) Assume that six reservations are made. Let X = the number of customers who have made a reservation and show up for the trip. Find the probability distribution function of X in table form.
2. B)Assume that six reservations are made. What is the probability that at least one individual with a reservation shows cannot be accommodated on the trip?
3. C)Assume that six reservations are made. What is the expected number of available places when the limousine departs?
4. Suppose that the probability distribution of the number of reservations made is given by the accompanying table. Let Y denote the number of passengers on a randomly selected trip.
5. D) Obtain the probability distribution function of Y in table form.The possible values for Y are 0, 1, 2, 3, 4. You are still assuming that 20% of passengers who have made a reservation do not show up.
6. E) Find the expected value of Y.

# of reservations	3	4	5	6
Probability	0.1	0.2	0.3	0.4

Answer 1

P(do not appear)=   0.2
P(appear) =   0.8

a)

available space ,Y	number of person appears	Probability
0	X≥4	P(X=4)+P(X=5)+P(X=6)=	0.9011
1	X=3	6C3*0.8^3*0.2^3=	0.0819
2	X=2	6C2*0.8^2*0.2^4=	0.0154
3	X=1	6C1*0.8^1*0.2^5=	0.0015
4	X=0	6C0*0.8^0*0.2^6=	0.0001

Y	P(Y)	Y*P(Y)
0	0.9011	0
1	0.0819	0.0819
2	0.0154	0.0307
3	0.0015	0.0046
4	0.0001	0.0003
total	1	0.1175

b).

P(at least one not accomodated) = P(X>4 appear) = P(X=5) + P(X=6)=   6C5*0.8^5*0.2^1+6C6*0.8^6=           0.6554

c)

The expected number of available places is   0.1175

d)

Number of reservations	3	4	5	6
Probability	0.1	0.2	0.3	0.4

Let Y shows the number of reservetion made. The X will take zero when no customer will show up.

X		Probability
0	P(X=0 and Y=3) + P(X=0 and Y=4) + P(X=0 and Y=5) + P(X=0 and Y=6)=	3C0*0.8^0*0.2^3*0.1 + 4C0*0.8^0*0.2^4*0.2+5C0*0.8^0*0.2^5*0.3 + 6C0*0.8^0*0.2^6*0.4=	0.0012
1	P(X=1 and Y=3) + P(X=1 and Y=4) + P(X=1 and Y=5) + P(X=1 and Y=6)=	3C1*0.8^1*0.2^2*0.1 + 4C1*0.8^1*0.2^3*0.2+5C1*0.8^1*0.2^4*0.3 + 6C1*0.8^1*0.2^5*0.4=	0.0173
2	P(X=2 and Y=3) + P(X=2 and Y=4) + P(X=2 and Y=5) + P(X=2 and Y=6)=	3C2*0.8^2*0.2^1*0.1 + 4C2*0.8^2*0.2^2*0.2+5C2*0.8^2*0.2^3*0.3 + 6C2*0.8^2*0.2^4*0.4=	0.0906
3	P(X=3 and Y=3) + P(X=3 and Y=4) + P(X=3 and Y=5) + P(X=3 and Y=6)=	3C3*0.8^3*0.2^0*0.1 + 4C3*0.8^3*0.2^1*0.2+5C3*0.8^3*0.2^2*0.3 + 6C3*0.8^3*0.2^3*0.4=	0.2273
4	1-P(X=0)-P(X=1)-P(X=2)-P(X=3)	1-0.0012-0.0173-0.0906-0.2273=	0.6636

X	0	1	2	3	4
P(X)	0.0012	0.0173	0.0906	0.2273	0.6636

E)

expected value = X*P(X) = 3.535