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

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: \({\rm{P(X}} \ge {\rm{5) = 0}}{\rm{.6553}}\).

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: \({\rm{E(g(X)) = 0}}{\rm{.117504}}\).

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