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, 20% 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.) If six reservations are made, what s the probability that at least one Individual with a reservation cannot be accommodated on the trip? 66 It six reservations are made, what s the expected number of available places when the limousine departs? 118 places Suppose the probability distribution of the number of reservations made is given in the accompanying table. Let X denote the number or passengers on a randomly selected trip. Obtain the probability mass function or X.

Explanation / Answer

P[X=0]= 3 reservation but none of them show up + 4 reservation but none of them show up+5 reservation but none of them show up + 6 reservation but none of them show up

= (0.14)*(0.2)^3+(0.22)*(0.2)^4+(0.34)*(0.2)^5+(0.3)*(0.2)^6

=0.0016 = 0.002 (round up to 3 decimal places)

P[X=1]= 0.8*[(0.14)*(0.2)^2+(0.22)*(0.2)^3+(0.34)*(0.2)^4+(0.3)*(0.2)^5]

=0.0064 = 0.006

P[X=2]=0.8^2*[(0.14)*(0.2)+(0.22)*(0.2)^2+(0.34)*(0.2)^3+(0.3)*(0.2)^4]

=0.0256 = 0.026

P[X=3]=0.8^3*[(0.14)*(0.2)^0+(0.22)*(0.2)^1+(0.34)*(0.2)^2+(0.3)*(0.2)^3]

=0.1024=0.102

P[X=4]=0.8^4*[(0.14)*0+(0.22)*(0.2)^0+(0.34)*(0.2)^1+(0.3)*(0.2)^2]

=0.12288= 0.123

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