An airport limousine can accommodate up to four passengers...

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, 40% of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate. (a) If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip? 0.23328 (b) If six reservations are made, what is the expected number of available places when the limousine departs? 0.6799 (c) Suppose the probability distribution of the number of reservations made is given in the accompanying table. Number of reservations Probability 3 0.09 4 0.25 5 0.32 6 0.34 Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X. X 0 1 2 3 4 p(x) 0.047 0.187 0.311 0.2765 0.1296

Transcript

00:01 In this question we have that an airport limousine can accommodate up to four passengers and the company will accept a maximum of six reservations and a passenger must have a reservation.

00:15 So the probability of a passenger not appearing for a reservation, so reserved, passenger, not appearing, we're told is 40%, 0 .40.

00:39 So if six reservations are made, what's the probability at least one individual with a reservation cannot be accommodated? so probability not everyone accommodated is equal to, so if six people make a reservation but only four people can fit in, then that means that either one or no people do not make their reservation, so either five or six people show up, which is probability zero or one don't show up.

01:23 So if we write that x is the number not showing up, then x is going to be binomial with six trials and a 0 .4 chance of success.

01:45 So let's call this probability p1.

01:50 P1 is going to be equal to the probability x equals zero plus the probability x equals one.

01:57 So the probability x equals zero is 0 .6 to the six and the probability x equals one is 0 .6 times 0 .4 times 0 .6 to the six, just by the definition of the binomial distribution.

02:14 Oh, sorry, that should be 0 .6 to the five.

02:21 And this gives me 0 .23328.

02:29 Part b, if six reservations are made, what is the expected number of available places? well, let's say that the number available is going to be the number of people that show up minus two.

02:44 Sorry, the number of people that don't show up minus two.

02:46 Because if two people don't show up, then everyone, you know, all of the places will be full.

02:52 And if more than two people don't show up, then there will be some free places.

02:56 So the expected number of available places is then the expected value of x minus two.

03:03 The expected value of x is just n times p.

03:09 So this is n, this is p.

03:11 N times p is six times 0 .4.

03:19 Six times 0 .4 minus two is 0 .4.

03:25 So that's the expected number of available places.

03:33 Now finally, the number of reservations and the probability that it takes that value are given by 3, 4, 5, or 6.

03:44 0 .09, 0 .25, 0 .32, 0 .34.

03:53 Now what we need is the pmf of the number of passengers.

03:57 So now x is now the number of passengers.

04:07 And x given nr is going to be binomial with nr reservations.

04:16 But the probability of turning up for a reservation is one minus the probability of not turning up.

04:22 So it's 0 .6.

04:29 So x, so the number of passengers on a trip...

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