A game of chance involves rolling a standard, six-sided die. The amount of money the player wins depends on the result of the die roll: * If the result is 1 or 2, the player wins nothing; * If the result is 3, 4, or 5, the player wins 4 dollars; * If the result is 6, the player wins 36 dollars. (Note: Your answer to the question below should be rounded to three decimal places.) If you play this game once (and it costs you nothing to play), what is the expected amount of money you will win?

Using the discrete probability principle, the expected value for the amount of money which could be won ls $8.00

X : $0 $4 $36

P(X) : __ 0.333 ___ 0.5 0.1667

The expected value ls defined thus :

E(X) = (0 × 0.333) + (4 × 0.5) + (36 × 1/6)

E(X) = 0 + 2.0 + 6

E(X) = $8.0

Therefore, the expected amount to be won in the long run is $8.0

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If you play this game once (and it costs you nothing to play), the expected amount of money you will win is:

$8.

Step-by-step explanation:

a) Data and Calculations:

The probability for the result to be 1 or 2 = 33.33% (100/6 * 2)

The probability for the result to be 3, 4, or 5 = 50% (100/6 * 3)

The probability for the result to be 6 = 16.67% (100/6 * 1)

Data Table:

Event                              Probability       Value Won

The result is 1 or 2           33.33%              $0

The result is 3, 4 or 5     50.00%              $4.00

The result is 6                 16.67%              $36.00

Expected Value:

Event                              Probability          Value              Expected Value

The result is 1 or 2           33.33%              $0                $0 ($0 * 33.33%)

The result is 3, 4 or 5     50.00%              $4.00           $2.00 ($4 * 50%)

The result is 6                 16.67%              $36.00          $6.00 ($36 * 16.67%)

Total expected amount of money  =                             $8.00

dude thats to long figure it

step-by-step explanation:

step-by-step explanation:

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