This is the problem:
Conside a small ferry that can accomidate cars and busses. The toll for cars is $3 and the toll for busses is $10. Let X and Y denote the number of cars and busses respectivly. Compute the expected revenue for the trip.
\(\displaystyle \begin{array}{{}}
{} & {} & {} & y & {} \\
{} & {p(x,y)} & 0 & 1 & 2 \\
{} & 0 & {.025} & {.015} & {.010} \\
{} & 1 & {.050} & {.030} & {.020} \\
x & 2 & {.125} & {.075} & {.050} \\
{} & 3 & {.150} & {.090} & {.060} \\
{} & 4 & {.100} & {.060} & {.040} \\
{} & 5 & {.050} & {.030} & {.020} \\
\end{array}\)
There were a bunch of easy questions asking the probability of certain situations which were easy, but the last one has me stumped (above)
I multiplied all of the x values by their probability sum and add the marginal totals to get x=2.8
Did the same with y and got y=.5
($10)(.5)+($3)(2.8)=$13.40
But the right anwser is $15.40, can anyone tell me where I'm going wrong? Thanks
Conside a small ferry that can accomidate cars and busses. The toll for cars is $3 and the toll for busses is $10. Let X and Y denote the number of cars and busses respectivly. Compute the expected revenue for the trip.
\(\displaystyle \begin{array}{{}}
{} & {} & {} & y & {} \\
{} & {p(x,y)} & 0 & 1 & 2 \\
{} & 0 & {.025} & {.015} & {.010} \\
{} & 1 & {.050} & {.030} & {.020} \\
x & 2 & {.125} & {.075} & {.050} \\
{} & 3 & {.150} & {.090} & {.060} \\
{} & 4 & {.100} & {.060} & {.040} \\
{} & 5 & {.050} & {.030} & {.020} \\
\end{array}\)
There were a bunch of easy questions asking the probability of certain situations which were easy, but the last one has me stumped (above)
I multiplied all of the x values by their probability sum and add the marginal totals to get x=2.8
Did the same with y and got y=.5
($10)(.5)+($3)(2.8)=$13.40
But the right anwser is $15.40, can anyone tell me where I'm going wrong? Thanks