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Justin is playing a game of chance in which he rolls a number cube with sides numbered from 1 to 6. The number cube is fair, so a side is rolled at random.This game is this: Justin rolls the number cube once. He wins $1 if a 1 is rolled, $2 if a 2 is rolled, $3 if a 3 is rolled, and $4 if a 4 is rolled. He loses $6.50 if a 5 or 6 is rolled.(If necessary, consult a list of formulas.)(a) Find the expected value of playing the game.dollars(b) What can Justin expect in the long run, after playing the game many times?Justin can expect to gain money.Hecanexpecttowindollarsperroll.Justin can expect to lose money.Hecanexpecttolosedollarsperroll.Justin can expect to break even (neither gain nor lose money).

Question

Justin is playing a game of chance in which he rolls a number cube with sides numbered from 1 to 6. The number cube is fair, so a side is rolled at random.This game is this: Justin rolls the number cube once. He wins 1ifa1isrolled,1 if a 1 is rolled, 2 if a 2 is rolled, 3ifa3isrolled,and3 if a 3 is rolled, and 4 if a 4 is rolled. He loses $6.50 if a 5 or 6 is rolled.(If necessary, consult a list of formulas.)(a) Find the expected value of playing the game.dollars(b) What can Justin expect in the long run, after playing the game many times?Justin can expect to gain money.Hecanexpecttowindollarsperroll.Justin can expect to lose money.Hecanexpecttolosedollarsperroll.Justin can expect to break even (neither gain nor lose money).

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Solution

To find the expected value of playing the game, we need to multiply each outcome by its probability and then sum these values.

The probability of rolling any number on a fair 6-sided die is 1/6.

So, the expected value (E) can be calculated as follows:

E = (1/6)1+(1/6)1 + (1/6)*2 + (1/6)3+(1/6)3 + (1/6)*4 - (1/6)*6.50(1/6)6.50 - (1/6)*6.50

E = 0.17+0.17 + 0.33 + 0.50+0.50 + 0.

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