91影视

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

Proposition: Number of events during a time interval of length\({\rm{t}}\)can be modelled using Poisson Random Variable with parameter\({\rm{\mu = \alpha t}}\). This indicates that 鈥�

\({{\rm{P}}_{\rm{k}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}\frac{{{{{\rm{(\alpha t)}}}^{\rm{k}}}}}{{{\rm{x!}}}}\)

It is also known as Poisson process (not formally defined).

A random variable\({\rm{X}}\)with\({\rm{pmf}}\)鈥�

\({\rm{p(x;\mu ) = }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\)

For\({\rm{x = 0,1,}}..{\rm{,}}\)is said to have Poisson Distribution with parameter\({\rm{\mu > 0}}\).

(a)

It is given that rate\({\rm{\alpha = 2}}\)per minute, and a\({\rm{1 - min}}\)period, therefore, there is a random variable\({\rm{X}}\)with Poisson Distribution with parameter 鈥�

\({\rm{\mu = \alpha t = 2}} \cdot {\rm{1 = 2}}\)

Appendix Table\({\rm{A}}{\rm{.2}}\)contains the Poisson cdf\({\rm{F(x;2)}}\).

The requested probability, for\({\rm{x = 0}}\), is 鈥�

\({\rm{P(X = 0) = F(0;2) = }}{{\rm{e}}^{{\rm{ - 2}}}}\frac{{{{\rm{2}}^{\rm{0}}}}}{{{\rm{0!}}}}{\rm{ = 0}}{\rm{.135}}\)

Therefore, the value obtained is \({\rm{0}}{\rm{.135}}\).

(b)

Denote with \({\rm{Y}}\) random variable 鈥�

\({\rm{Y = }}\)the number of operators that receive no requests.

Such random variable follows Binomial Distribution with parameters \({\rm{n = 5}}\) and \({\rm{p = P(}}\)operators that receive no requests\({\rm{) = 0}}{\rm{.135}}\).

Which is calculated in (a) because \({\rm{1 - min}}\) period is given. Therefore,

\({\rm{Y}} \sim {\rm{Bin(5,0}}{\rm{.135)}}\)

For the \({\rm{pmf}}\) of Binomial Distribution the following theorem holds.

Theorem:

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

Hence, the probability that four out of five operators receive no requests is 鈥�

\(\begin{array}{c}{\rm{P(Y = 4) = b(4;5,0,0}}{\rm{.135)}}\\{\rm{ = }}\left( {\begin{array}{*{20}{c}}{\rm{5}}\\{\rm{4}}\end{array}} \right){\rm{0}}{\rm{.13}}{{\rm{5}}^{\rm{4}}}{{\rm{(1 - 0}}{\rm{.0}}{\rm{.135)}}^{{\rm{5 - 4}}}}\\{\rm{ = 0}}{\rm{.0014}}\end{array}\)

Therefore, the value obtained is \({\rm{0}}{\rm{.0014}}\).

(c)

Note that the operators receive requests independently of one another. Denote event 鈥�

\({{\rm{A}}_{{\rm{ik}}}}{\rm{ = }}\){\({{\rm{i}}^{{\rm{th}}}}\)operator receives exactly\({\rm{k}}\)requests}

Let \({\rm{k}} \ge {\rm{0}}\), the following is true 鈥�

\(\begin{array}{c}{\rm{P}}\left( { \cap _{{\rm{i = 1}}}^{\rm{5}}{{\rm{A}}_{{\rm{ik}}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{{\rm{1k}}}}} \right) \cdot {\rm{P}}\left( {{{\rm{A}}_{{\rm{2k}}}}} \right) \cdot {\rm{ \ldots }} \cdot {\rm{P}}\left( {{{\rm{A}}_{{\rm{5k}}}}} \right)\\{\rm{ = p(k;2}}{{\rm{)}}^{\rm{5}}}{\rm{ = }}{\left( {{{\rm{e}}^{{\rm{ - 2}}}}\frac{{{{\rm{2}}^{\rm{k}}}}}{{{\rm{k!}}}}} \right)^{\rm{5}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - 10}}}}\frac{{{{\rm{2}}^{{\rm{4k}}}}}}{{{\rm{(k!)}}}}\end{array}\)

Since number \({\rm{k}}\) was non negative integer, the probability that all receive exactly the same number of requests would be union of disjoint events 鈥�

\({{\rm{B}}_{\rm{k}}}{\rm{ = }} \cap _{{\rm{i = 1}}}^{\rm{5}}{{\rm{A}}_{{\rm{ik}}}}{\rm{,}}\;\;\;{\rm{k}} \ge {\rm{0}}\)

Therefore, if it is denoted that 鈥�

\({\rm{B = }}\)all operators receive the same number of requests.

Then it is obtained 鈥�

\(\begin{array}{c}{\rm{P(B) = P}}\left( { \cup _{{\rm{k = 0}}}^\infty {{\rm{B}}_{\rm{k}}}} \right){\rm{ = }}\sum\limits_{k = 0}^\infty {P({B_k})} \\{\rm{ = }}\sum\limits_{{\rm{k = 0}}}^\infty {{{\rm{e}}^{{\rm{ - 10}}}}} \frac{{{{\rm{2}}^{{\rm{5k}}}}}}{{{{{\rm{(k!)}}}^{\rm{5}}}}}\end{array}\)

Therefore, the expression is obtained as \(\sum\limits_{{\rm{k = 0}}}^\infty {{{\rm{e}}^{{\rm{ - 10}}}}} \frac{{{{\rm{2}}^{{\rm{5k}}}}}}{{{{{\rm{(k!)}}}^{\rm{5}}}}}\).