document.write( "Question 1205333: In a raffle where 3500 tickets are sold for $2 each, one prize of $4800 will be awarded. What is the expected value of a single ticket in the raffle?
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Algebra.Com's Answer #842064 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Let's say one single person buys all 3500 tickets.\r
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\n" ); document.write( "\n" ); document.write( "They would spend 3500*2 = 7000 dollars in total.
\n" ); document.write( "They would have 100% probability of getting the prize of $4800.
\n" ); document.write( "But they walk away with a net loss of 4800 - 7000 = -2200 dollars.\r
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\n" ); document.write( "\n" ); document.write( "Divide this net loss over the number of tickets.
\n" ); document.write( "-2200/3500 = -0.62857 approximately
\n" ); document.write( "This rounds to -0.63 which is the approximate expected value.\r
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\n" ); document.write( "\n" ); document.write( "The person should expect to lose, on average, around 63 cents per ticket.\r
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\n" ); document.write( "\n" ); document.write( "Another approach is to set up a table as shown below
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EventXP(X)X*P(X)
Win47981/35004798/3500
Lose-23499/3500-6998/3500

\n" ); document.write( "where X = net earnings = amount you walk away with
\n" ); document.write( "Note that 4800 - 2 = 4798
\n" ); document.write( "P(X) = probability of getting those earnings\r
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\n" ); document.write( "\n" ); document.write( "The X*P(X) column is the result of multiplying each X and P(X) value together. \r
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\n" ); document.write( "\n" ); document.write( "Add up those X*P(X) values:
\n" ); document.write( "4798/3500 + (-6998/3500)
\n" ); document.write( "4798/3500 - 6998/3500
\n" ); document.write( "(4798 - 6998)/3500
\n" ); document.write( "-2200/3500
\n" ); document.write( "-0.62857 approximately
\n" ); document.write( "That leads to -0.63 when rounding to the nearest cent.\r
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\n" ); document.write( "\n" ); document.write( "Answer: -0.63 dollars
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