Agame is played using one die. if the die is rolled and shows 5 5, the player wins $ 10 $10. if the die shows any number other than 5 5, the player wins nothing. if there is a charge of $ 2 $2 to play the game, what is the game's expected value?

The expected value is -$0.33

Step-by-step explanation: The expected value is calculated as:

E = ∑pₙxₙ

where xₙ is a given event (in this case winning 8$ or losing 2$) and pₙ is the probability for the event.

So in this you only win with one result in the dice, so the probability of winning is 1/6, and the probability of losing must be 5/6.

Then the expected value is:

E = 1/6*$8 - 5/6*$2 = $-0.33

Where I used the profit of $8 instead of $10, because you must pay $2 for play, and you win $10, the profit is ($10 - $2) = $8

the solution is h=10 because you add 9 to each side of the equation.

answer: add the !

step-by-step explanation: then we will answer