Question

pschoolsri.com bookmarks A No-Nonsense Nu.ServicePortal(Summ2022) MATH-1139-601 Math for Liberal Arts Students DLHomework: Section 11.8Kara Roberson ParChoose the correct answer below, and it in the answer box to complete your choice.(Round to the nearest cent as needed. Do not include the $ symbol in your answer)Help me solve thisKara Roberson TView an example Get more help.Question 4, 11.8.27A. The statement does not make sense because the expected value after ten rots is dollars, which is greater than the value of the current billOB. The statement makes sense because the expected value after ten rots is dollars, which is less than the value of the current billFREE RST TRAINNhpMy Account SettingsAdvisory-ThibeaulDetermine whether the following statement makes sense or does not make sense, and explain your reasoningHere's my dilemma, I can accept a $1400 bill or play a game ten times. For each roll of the single die, I win $700 for rolling 1 or 2; I win $200 for rolling 3, and I lose $600 for rolling 4, 5, or 6. Based on the expected value, I shouldaccept the $1400 billkara Roberson 06/22/22 12:25 PMHW Score: 75%, 3 of 4 pointsO Points: 0 of 1Reading listClear allSaveCheck answer12:26

Answer 1

Answer

B. The statement makes sense because the expected value after ten rolls is -333.33 dollars, which is less than the value of the current bill

Step-by-step explanation

The expected value per bet is calculated as follows:

$\text{ Expected value \lparen per bet\rparen }=p(winning)\times amount\text{ won}-p(losing)\times amount\text{ lost}$

There are 6 outcomes when rolling a die, then the probability of rolling a 1 or a 2, the probability of rolling a 3, and the probability of rolling a 4 or a 5 or a 6 is:

$\begin{gathered} p(rolling\text{ 1 or 2})=\frac{2}{6}=\frac{1}{3} \\ p(roll\imaginaryI ng\text{ 3})=\frac{1}{6} \\ p(roll\imaginaryI ng\text{ 4 or 5 or 6})=\frac{3}{6}=\frac{1}{2} \end{gathered}$

You win $700 for rolling 1 or 2, $200 for rolling 3, and you lose $600for rolling 4, 5, or 6, then the expected value per bet is:

$\begin{gathered} \text{ Expected value \lparen per bet\rparen }=\frac{1}{3}\times\text{ \ }700+\frac{1}{6}\times\text{ \ 2}00-\frac{1}{2}\times\text{ \ 6}00 \\ \text{Expected value }\operatorname{\lparen}\text{per bet}\operatorname{\rparen}=-\text{ \ }33.33 \end{gathered}$

You have to expect to lose $33.33 per bet. After 10 bets, the expected value is:

$\begin{gathered} \text{ Expected value = number of bets}\times Expected\text{ }value\text{ per bet} \\ Expected\text{ }value=10\times-\text{ \ }33.33 \\ Expected\text{ v}alue=-\text{ \ }333.33 \end{gathered}$