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Calculate the Expected Monetary Value (EMV), in thousands of pounds, for each option:
\n\n | EMV | \n
{Cat[0]} | \n[[0]] | \n
{Cat[1]} | \n[[1]] | \n
{Cat[2]} | \n[[2]] | \n
Hence determine the optimal course of action:
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Given data on probabilities of three levels of success of three options and projections of the profits that the options will accrue depending on the level of success, find the expected monetary value (EMV) for each option and choose the one with the greatest EMV.
"}, "tags": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "advice": "For a given option, we can find the Expected Monetary Value by:
\ne.g. the Expected Monetary Value for \"{Cat[0]}\" is given in four steps (all numbers below are in {units}):
\n1. Multiplying the probability $\\var{p1}$ of a {Att[1]} outcome by the expected profit $\\var{a[0][0]}$, gives:
\nexpected profit = $\\var{p1}\\times \\var{a[0][0]}= \\var{p1*a[0][0]}$
\n\n
2. Multiplying the probability $\\var{p2}$ of a {Att[2]} outcome by the expected profit, $\\var{a[0][1]}$ gives:
\nexpected profit = $\\var{p2}\\times \\var{a[0][1]}= \\var{p2*a[0][1]}$
\n\n
3. Multiplying the probability $\\var{p3}$ of a {Att[3]} outcome by the expected profit, $\\var{a[0][2]}$ gives:
\nexpected profit = $\\var{p3}\\times \\var{a[0][2]}= \\var{p3*a[0][2]}$
\n4. Finally add these three together to get the Expected Monetary Value for the option {Cat[0]} :
\n$\\var{p1*a[0][0]}+\\var{p2*a[0][1]}-\\var{abs(p3*a[0][2])}=\\var{emv[0]}$
\n\n
You calculate in the same way for the other options - the next table gives the Expected Monetary Value for all three::
\n\n | EMV | \n
{Cat[0]} | \n{emv[0]} | \n
{Cat[1]} | \n{emv[1]} | \n
{Cat[2]} | \n{emv[2]} | \n
The optimal course of action is take to be that which has the highest Expected Monetary Value (EMV) and this is seen to be :
\n{Correct} with EMV $\\var{maxemv}$.
\n\n
\n
\n
\n
", "statement": "
{Something} {hasdonethis} {decision}
\nA. {Cat[0]}
\nB. {Cat[1]}
\nC. {Cat[2]}
\n{info} has the following probabilities associated to the following {outcomes} for the {product}:
\n{table([['<strong>Probability:</strong>']+p],Att)}
\nThe next table shows {expectedreturn} for each option against these {outcomes}:
\n{table(b,Att)}
", "type": "question", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}