An airport limousine can accommodate up to four passengers on any one trip. The

Question

An airport limousine can accommodate up to four passengers on any one trip. The company will accept a maximum of six reservations for a trip, and a passenger must have a reservation. From previous records, 10% of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate. (Round your answers to three decimal places.)

Suppose the probability distribution of the number of reservations made is given in the accompanying table.

Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X.

Number of reservations     3 4 5 6 Probability   0.14        0.23        0.30        0.33     

Explanation / Answer

Pr(ANy reservations will not appear for the trip) = 0.10

Pr(any reservation will appear for the trip) = 0.90

Here if number of reservations (lets say y) follow the given distribution.

Here X denote the number of passengers on a randomly selected trip.

Here,

P(x) = Pr(Y = 3) * Pr(X l Y = 3) + Pr(Y = 4) * Pr(X l Y = 4) + Pr(Y = 5) * Pr(X l Y = 5) +Pr(Y = 6) * Pr(X l Y = 6)

so,

P(x = 0) = 0.14 * BIN(X = 0; 3; 0.90) + 0.23 * BIN(X = 0 ; 4; 0.90) + 0.30 * BIN(X = 0; 5; 0.90) + 0.33 * BIN(X = 0; 6; 0.90)

= 0.14 * 0.001 + 0.23* 10-5 + 0.30 * 10-5 + 0.33 * 10-6 = 0.000166

Pr(x = 1) = 0.14 * BIN(X = 1; 3; 0.90) + 0.23 * BIN(X = 1 ; 4; 0.90) + 0.30 * BIN(X = 1; 5; 0.90) + 0.33 * BIN(X = 1; 6; 0.90)

= 0.14 * 0.027 + 0.23 * 0.0036 + 0.30 * 0.00045 + 0.33 * 5.4 * 10-5 = 0.004761

Pr(x = 2) = 0.14 * BIN(X = 2; 3; 0.90) + 0.23 * BIN(X = 2 ; 4; 0.90) + 0.30 * BIN(X =2; 5; 0.90) + 0.33 * BIN(X = 2; 6; 0.90)

= 0.14 * 0.243 + 0.23 * 0.0486 + 0.30 * 0.0081 + 0.33 * 0.001215 = 0.04803

Pr(x = 3)   = 0.14 * BIN(X = 3; 3; 0.90) + 0.23 * BIN(X = 3 ; 4; 0.90) + 0.30 * BIN(X =3; 5; 0.90) + 0.33 * BIN(X = 3; 6; 0.90)

= 0.14 * 0.729 + 0.23 * 0.2916 + 0.30 * 0.0729 + 0.33 * 0.01458 = 0.1958

Pr(x = 4) = 1 - [Pr(x=0) + Pr( x =1) + Pr(x =2) + Pr(x =3) + Pr(x = 4)] = 0.7512

so the probability mass function tabe is

X P(x) 0 0.000166 1 0.004761 2 0.048029 3 0.195809 4 0.751235

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