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,\({\rm{20 \% }}\)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? b. If six reservations are made, what is the expected number of available places when the limousine departs? c. 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.

Short answer

(a) The probability that at least one individual with a reservation cannot be accommodated on a trip is obtained as: \({\rm{P(X}} \ge {\rm{5) = 0}}{\rm{.6553}}\).

(b) The expected number of available places when limousine departs is obtained as: \({\rm{E(g(X)) = 0}}{\rm{.117504}}\).

(c) The probability mass function of X is obtained as:

\(p(x) = \left\{ {\begin{array}{*{20}{l}}{0.00124}&{,x = 0} \\ {0.01725}&{,x = 1} \\{0.09062}&{,x = 2} \\ {0.22733}&{,x = 3} \\ {0.66355}&{,x = 4} \\ 0&{,x\ddot I \{ 0,1, \ldots ,4\} } \end{array}} \right.\)

Step-by-step solution

Define Discrete random variables

A discrete random variable is one that can only take on a finite number of different values.

Step 2: What is the probability that at least one individual with a reservation cannot be accommodated on the trip?

(a) X stands for the number of bookings that show.

The random variable X has a Binomial Distribution with the following parameters: \({\rm{n = 6}}\) (six reservations are made) and \(\begin{array}{c}{\rm{p = 1 - 0}}{\rm{.2}}\\{\rm{ = 0}}{\rm{.8}}\end{array}\) (since \({\rm{20\% }}\) do not show, indicating that \({\rm{80 \% }}\) do). So,

\({\rm{X}} \sim {\rm{Bin(6,0}}{\rm{.8)}}\)

The chance that we must compute is (since the limousine can seat up to four people, thus if five or six arrive, at least one will be unable to be accommodated).

\(\begin{gathered}P(X \geqslant 5) = 1 - P(X < 5) \\ \mathop = \limits^{(1)} 1 - P(X \leqslant 4) \\ \mathop = \limits^{(2)} 1 - B(4;6,0.8) \\ \mathop = \limits^{(3)} 1 - 0.3446 \\ = 0.6553 \\ \end{gathered} \)

(1): X can only handle non-negative integers \({\rm{3}}\);

(2): See the Binomial Distribution cdf at the bottom of the page;

(3):You may get it from Appendix Table A.\({\rm{1}}\). or you can do it yourself.

The cdf of a binomial random variable X with parameters n and p is the Cumulative Density Function.

\(\begin{array}{c}{\rm{B(x;n,p) = P(X < 5)}}\\{\rm{ = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {{\rm{b(y;n,p)}}} \\{\rm{x = 0,1,}}....{\rm{,n}}\end{array}\)

Therefore, the probability is: \({\rm{P(X}} \ge {\rm{5) = 0}}{\rm{.6553}}\).

Step 3: What is the expected number of available places when the limousine departs?

(b) We'll stick with the same random variable X from part one (a). The number of available spots will be determined by the number of persons who show up; for example, if only six people show up, no spots will be available.

Similarly, if \({\rm{0}}\) people show up, \({\rm{4}}\) (all) available spots will be filled.

As a result, use g(X) to signify the newly produced random variable as follows:

T

The Expected Value (mean value) of any function g(X), where X is a discrete random variable X with a set of potential values S and pmf p(x), indicated as E(g(X)) \({\rm{(}}{{\rm{\mu }}_{{\rm{g(x)}}}}{\rm{)}}\).

\(\begin{array}{c}{\rm{E(g(x)) = (}}{{\rm{\mu }}_{{\rm{g(x)}}}}{\rm{)}}\\{\rm{ = }}\sum\limits_{{\rm{x}} \in {\rm{s}}} {{\rm{g(x) \bullet p(x)}}} \end{array}\)

The following is true based on the above proposition:

\(\begin{array}{c}{\rm{E(g(X)) = }}\sum\limits_{{\rm{x = 0}}}^{\rm{6}} {\rm{g}} {\rm{(x) \times p(x)}}\\{\rm{ = }}\sum\limits_{{\rm{x = 0}}}^{\rm{6}} {\rm{g}} {\rm{(x) \times b(x;6,0}}{\rm{.8)}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{4 \times b(0;6,0}}{\rm{.8) + 3 \times b(1;6,0}}{\rm{.8) + 2 \times b(2;6,0}}{\rm{.8) + 1 \times b(3;6,0}}{\rm{.8)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{4 \times }}\left( {\begin{array}{*{20}{l}}{\rm{6}}\\{\rm{0}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{8}}^{\rm{0}}}{{\rm{(1 - 0}}{\rm{.8)}}^{{\rm{6 - 0}}}}{\rm{ + 3 \times }}\left( {\begin{array}{*{20}{l}}{\rm{6}}\\{\rm{1}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{8}}^{\rm{1}}}{{\rm{(1 - 0}}{\rm{.8)}}^{{\rm{6 - 1}}}}{\rm{ + 2 \times }}\left( {\begin{array}{*{20}{l}}{\rm{6}}\\{\rm{2}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{8}}^{\rm{2}}}{{\rm{(1 - 0}}{\rm{.8)}}^{{\rm{6 - 2}}}}{\rm{ + 1 \times }}\left( {\begin{array}{*{20}{l}}{\rm{6}}\\{\rm{3}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{8}}^{\rm{3}}}{{\rm{(1 - 0}}{\rm{.8)}}^{{\rm{6 - 3}}}}\\{\rm{ = 4 \times 0}}{\rm{.00006 + 3 \times 0}}{\rm{.00154 + 2 \times 0}}{\rm{.01536 + 1 \times 0}}{\rm{.08192}}\\{\rm{ = 0}}{\rm{.117504}}\end{array}\)

(1):As, g(x) =\({\rm{0}}\), all the rest are zero;

(2): observe the following theorem.

Theorem:

\({\rm{b(x;n,p) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}&{{\rm{,x = 0,1,2, \ldots ,n}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

Therefore, the expected number is: \({\rm{E(g(X)) = 0}}{\rm{.117504}}\).

Step 4: Obtaining the probability mass function of X.

(c) The number of passengers cannot exceed four, and only four persons will be allowed to participate in the specified trip. Reservations might range from three to six.

As a result, if no one shows up for the randomly picked journey, there will be no passengers. This can happen in four scenarios.

\({\rm{3}}\)reservations with no one showing up; \({\rm{4}}\) reservations with no one showing up; \({\rm{5}}\) reservations with no one showing up; \({\rm{6}}\) reservations with no one showing up. Indicate the occurrences.

\(\begin{array}{c}{{\rm{R}}_{\rm{i}}}{\rm{ = \{ i passenger reservations\} ;}}\\{{\rm{S}}_{\rm{i}}}{\rm{ = \{ i passenger show up\} }}{\rm{.}}\end{array}\)

As a result, the following is correct:

\(\begin{array}{c}{\rm{P\{ X = 0\} }}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{0}}}} \right) \cup \left( {{{\rm{R}}_{\rm{4}}} \cap {{\rm{S}}_{\rm{0}}}} \right) \cup \left( {{{\rm{R}}_{\rm{5}}} \cap {{\rm{S}}_{\rm{0}}}} \right) \cup \left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{0}}}} \right)} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P}}\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{0}}}} \right){\rm{ + \ldots + }}\left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{0}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{P}}\left( {{{\rm{S}}_{\rm{0}}}\mid {{\rm{R}}_{\rm{3}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{3}}}} \right){\rm{ + \ldots + P}}\left( {{{\rm{S}}_{\rm{0}}}\mid {{\rm{R}}_{\rm{6}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{6}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{b(0;3,0}}{\rm{.8) \times 0}}{\rm{.1 + b(0;4,0}}{\rm{.8) \times 0}}{\rm{.2 + b(0;5,0}}{\rm{.8) \times 0}}{\rm{.3 + b(0;6,0}}{\rm{.8) \times 0}}{\rm{.4}}\\\mathop {\rm{ = }}\limits^{{\rm{(5)}}} {\rm{0}}{\rm{.008 \times 0}}{\rm{.1 + 0}}{\rm{.0016 \times 0}}{\rm{.2 + 0}}{\rm{.00032 \times 0}}{\rm{.3 + 0}}{\rm{.000064 \times 0}}{\rm{.4}}\\{\rm{ = 0}}{\rm{.00124}}\end{array}\)

(1):as previously stated, the event \({\rm{\{ X = 0\} }}\) may be modelled as the union of four discrete events;

(2):the sequence of events is disjointed;

(3) the following multiplication rule;

(4):in the exercise, the probability of occurrences \({{\rm{R}}_{\rm{i}}}\) are provided in a table; the likelihoods of happenings \({{\rm{S}}_{\rm{0}}}{\rm{|}}{{\rm{R}}_{\rm{i}}}\) may be determined using the binomial distribution (the number of trials varies based on the number of reservations!).

(5): the values may be computed using the same theorem as in (b).

The Rule of Multiplication

\({\rm{P(A}} \cap {\rm{B) = P(A|B) \bullet P(B)}}\)

Similarly, for the occurrence \({\rm{\{ X = 1\} }}\), which means that just one passenger goes on a journey, we have:

\(\begin{array}{c}{\rm{P\{ X = 1\} }}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{1}}}} \right) \cup \left( {{{\rm{R}}_{\rm{4}}} \cap {{\rm{S}}_{\rm{1}}}} \right) \cup \left( {{{\rm{R}}_{\rm{5}}} \cap {{\rm{S}}_{\rm{1}}}} \right) \cup \left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{1}}}} \right)} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P}}\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{1}}}} \right){\rm{ + \ldots + }}\left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{1}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{P}}\left( {{{\rm{S}}_{\rm{1}}}\mid {{\rm{R}}_{\rm{3}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{3}}}} \right){\rm{ + \ldots + P}}\left( {{{\rm{S}}_{\rm{1}}}\mid {{\rm{R}}_{\rm{6}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{6}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{b(1;3,0}}{\rm{.8) \times 0}}{\rm{.1 + b(1;4,0}}{\rm{.8) \times 0}}{\rm{.2 + b(1;5,0}}{\rm{.8) \times 0}}{\rm{.3 + b(1;6,0}}{\rm{.8) \times 0}}{\rm{.4}}\\\mathop {\rm{ = }}\limits^{{\rm{(5)}}} {\rm{0}}{\rm{.096 \times 0}}{\rm{.1 + 0}}{\rm{.0256 \times 0}}{\rm{.2 + 0}}{\rm{.0064 \times 0}}{\rm{.3 + 0}}{\rm{.001536 \times 0}}{\rm{.4}}\\{\rm{ = 0}}{\rm{.01725}}\end{array}\)

(1): As previously stated, the event \({\rm{\{ X = 1\} }}\) may be modelled as the union of four discrete occurrences.

(2):the sequence of events is disjointed;

(3) the above-mentioned multiplication rule;

(4):in the exercise, the probability of occurrences \({{\rm{R}}_{\rm{i}}}\) are provided in a table; the probability of \({{\rm{S}}_{\rm{1}}}{\rm{|}}{{\rm{R}}_{\rm{i}}}\) events; may be computed using the binomial distribution (the number of trials fluctuates based on the number of reservations! );

(5): the values may be computed using the same theorem as in (b).

Similarly, for the occurrence \({\rm{\{ X = 2\} }}\), which means that just one passenger goes on a journey, we have:

\(\begin{array}{c}{\rm{P\{ X = 2\} }}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{2}}}} \right) \cup \left( {{{\rm{R}}_{\rm{4}}} \cap {{\rm{S}}_{\rm{2}}}} \right) \cup \left( {{{\rm{R}}_{\rm{5}}} \cap {{\rm{S}}_{\rm{2}}}} \right) \cup \left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{2}}}} \right)} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P}}\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{2}}}} \right){\rm{ + \ldots + }}\left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{2}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{P}}\left( {{{\rm{S}}_{\rm{2}}}\mid {{\rm{R}}_{\rm{3}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{3}}}} \right){\rm{ + \ldots + P}}\left( {{{\rm{S}}_{\rm{2}}}\mid {{\rm{R}}_{\rm{6}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{6}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{b(2;3,0}}{\rm{.8) \times 0}}{\rm{.1 + b(2;4,0}}{\rm{.8) \times 0}}{\rm{.2 + b(2;5,0}}{\rm{.8) \times 0}}{\rm{.3 + b(2;6,0}}{\rm{.8) \times 0}}{\rm{.4}}\\\mathop {\rm{ = }}\limits^{{\rm{(5)}}} {\rm{0}}{\rm{.384 \times 0}}{\rm{.1 + 0}}{\rm{.1536 \times 0}}{\rm{.2 + 0}}{\rm{.0512 \times 0}}{\rm{.3 + 0}}{\rm{.01536 \times 0}}{\rm{.4}}\\{\rm{ = 0}}{\rm{.09062}}\end{array}\)

(1): As previously stated, the event \({\rm{\{ X = 2\} }}\) may be modelled as the union of four discrete occurrences.

(2):the sequence of events is disjointed;

(3) the above-mentioned multiplication rule;

(4):in the exercise, the probability of occurrences \({{\rm{R}}_{\rm{i}}}\) are provided in a table; the probability of \({{\rm{S}}_{\rm{2}}}{\rm{|}}{{\rm{R}}_{\rm{i}}}\) events; may be computed using the binomial distribution (the number of trials fluctuates based on the number of reservations! );

(5): the values may be computed using the same theorem as in (b).

Similarly, for the occurrence \({\rm{\{ X = 3\} }}\), which means that just one passenger goes on a journey, we have:

\(\begin{array}{c}{\rm{P\{ X = 3\} }}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{3}}}} \right) \cup \left( {{{\rm{R}}_{\rm{4}}} \cap {{\rm{S}}_{\rm{3}}}} \right) \cup \left( {{{\rm{R}}_{\rm{5}}} \cap {{\rm{S}}_{\rm{3}}}} \right) \cup \left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{3}}}} \right)} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P}}\left( {{{\rm{R}}_{\rm{3}}} \cap {{\rm{S}}_{\rm{3}}}} \right){\rm{ + \ldots + }}\left( {{{\rm{R}}_{\rm{6}}} \cap {{\rm{S}}_{\rm{3}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{P}}\left( {{{\rm{S}}_{\rm{3}}}\mid {{\rm{R}}_{\rm{3}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{3}}}} \right){\rm{ + \ldots + P}}\left( {{{\rm{S}}_{\rm{3}}}\mid {{\rm{R}}_{\rm{6}}}} \right){\rm{ \times P}}\left( {{{\rm{R}}_{\rm{6}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{b(3;3,0}}{\rm{.8) \times 0}}{\rm{.1 + b(3;4,0}}{\rm{.8) \times 0}}{\rm{.2 + b(3;5,0}}{\rm{.8) \times 0}}{\rm{.3 + b(3;6,0}}{\rm{.8) \times 0}}{\rm{.4}}\\\mathop {\rm{ = }}\limits^{{\rm{(5)}}} {\rm{0}}{\rm{.512 \times 0}}{\rm{.1 + 0}}{\rm{.4096 \times 0}}{\rm{.2 + 0}}{\rm{.2048 \times 0}}{\rm{.3 + 0}}{\rm{.08192 \times 0}}{\rm{.4}}\\{\rm{ = 0}}{\rm{.22733}}\end{array}\)

(1): As previously stated, the event \({\rm{\{ X = 3\} }}\) may be modelled as the union of four discrete occurrences.

(2):the sequence of events is disjointed;

(3) the above-mentioned multiplication rule;

(4):in the exercise, the probability of occurrences \({{\rm{R}}_{\rm{i}}}\) are provided in a table; the probability of \({{\rm{S}}_{\rm{3}}}{\rm{|}}{{\rm{R}}_{\rm{i}}}\) events; may be computed using the binomial distribution (the number of trials fluctuates based on the number of reservations! );

(5): the values may be computed using the same theorem as in (b).

Now, for \({\rm{\{ X = 4\} }}\)the following is true:

\(\begin{array}{c}{\rm{P\{ X = 4\} = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)) }}\\{\rm{ = 1 - 0}}{\rm{.00124 - 0}}{\rm{.01725 - 0}}{\rm{.09062 - 0}}{\rm{.22733}}\\{\rm{ = 0}}{\rm{.66355}}\end{array}\)

As, the random variable can only take the values \({\rm{0, 1, 2, 3}}\), and \({\rm{4}}\), the pmf of X is zero for all other \({\rm{x }} \in {\rm{ R}}\).

Finally, in the exercise, the pmf of random variable X is:

\({\rm{p(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.00124}}}&{{\rm{,x = 0}}}\\{{\rm{0}}{\rm{.01725}}}&{{\rm{,x = 1}}}\\{{\rm{0}}{\rm{.09062}}}&{{\rm{,x = 2}}}\\{{\rm{0}}{\rm{.22733}}}&{{\rm{,x = 3}}}\\{{\rm{0}}{\rm{.66355}}}&{{\rm{,x = 4}}}\\{\rm{0}}&{{\rm{,x\ddot I \{ 0,1, \ldots ,4\} }}}\end{array}} \right.\)

Therefore, the mass function of X is: \({\rm{p(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.00124}}}&{{\rm{,x = 0}}}\\{{\rm{0}}{\rm{.01725}}}&{{\rm{,x = 1}}}\\{{\rm{0}}{\rm{.09062}}}&{{\rm{,x = 2}}}\\{{\rm{0}}{\rm{.22733}}}&{{\rm{,x = 3}}}\\{{\rm{0}}{\rm{.66355}}}&{{\rm{,x = 4}}}\\{\rm{0}}&{{\rm{,x\ddot I \{ 0,1, \ldots ,4\} }}}\end{array}} \right.\).