SOLUTION
The expected value can be calculated using the formula
Probability of selecting tickets worth $2000 =
[tex]\begin{gathered} \frac{1}{700} \\ \sin ce\text{ there is only one ticket which worth 2000 dollars } \end{gathered}[/tex]
Probability of selecting tickets worth $500 and
that of $40 are
[tex]\frac{2}{700}\text{ and }\frac{11}{700}\text{ respectively }[/tex]
[tex]\begin{gathered} \frac{1}{700} \\ \sin ce\text{ there is only one ticket which worth 2000 dollars } \end{gathered}[/tex][tex]\begin{gathered} \Sigma_xP(x)=2000(\frac{1}{700})+500(\frac{2}{700})+40(\frac{11}{700})-\cos t\text{ of a ticket } \\ \Sigma_xP(x)=2000(\frac{1}{700})+500(\frac{2}{700})+40(\frac{11}{700})-10 \end{gathered}[/tex]
This becomes
[tex]\begin{gathered} \Sigma_xP(x)=2000(\frac{1}{700})+500(\frac{2}{700})+40(\frac{11}{700})-10 \\ \Sigma_xP(x)=\frac{20}{7}+\frac{10}{7}+\frac{22}{35}-10 \\ \Sigma_xP(x)=-\frac{178}{35} \\ \Sigma_xP(x)=-5.0857 \\ \Sigma_xP(x)=-5.09 \end{gathered}[/tex]
Hence the answer is $-5.09 to the nearest cent.
