91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

The

Expected Value

(Mean value) of a random variable\({\rm{g(X,Y)}}\), where \({\rm{g( \times )}}\) is a function, denoted as \({\rm{E(g(X,Y))}}\) is given by

\(E(g(X,Y)) = \left\{ {\begin{aligned}{*{20}{l}}{\sum\limits_x {\sum\limits_y g } (x,y) \cdot p(x,y)}&{,X{\rm{ and }}Y{\rm{ discrete }}}\\{\int_{ - \infty }^\infty {\int_{ - \infty }^\infty g } (x,y) \cdot f(x,y)dxdy}&{,X{\rm{ and }}Y{\rm{ continuous}}{\rm{. }}}\end{aligned}} \right.\)

where \({\rm{p(x,y)}}\) is pmf and \({\rm{f(x,y)}}\) pdf.

We are given pmf, therefore we can compute the expected revenue from a single trip, where \({\rm{g(X,Y) = 3X + 10Y(3}}\) dollars for cars and \({\rm{10}}\) dollars for buses). The following is true

\(\begin{aligned}{\rm{E(3X + 10Y) = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {{\rm{(3x + 10y)}}} } {\rm{ \times p(x,y)}}\\{\rm{ = (3 \times 0 + 10 \times 0) \times p(0,0) + (3 \times 0 + 10 \times 1) \times p(0,1) + \ldots + (3 \times 5 + 10 \times 2) \times p(5,2)}}\\{\rm{ = 0 \times 0}}{\rm{.025 + 10 \times 0}}{\rm{.015 + \ldots + 35 \times 0}}{\rm{.02}}\\{\rm{ = 15}}{\rm{.4}}{\rm{.}}\end{aligned}\)

Therefore, the expected revenue is\({\rm{E(3X + 10Y) = 15}}{\rm{.4}}\).