Dvice a game of chance using a fair die with corresponding winnings and losses For example, in rolling a die who me a sign when outcomes as winning outcomes and the odd outcomes as the losses or team of your own plan. Let x the amount to be won by a player or the amount he will lost when he plays your game .Compute for the mean or the average amount a player may win or lose the variance and the standard deviation and the pressure results to forget that each phase of a die has a probability of 1/6.
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Step-by-step explanation:
One possible game of chance using a fair die is as follows:
-The player rolls the die.
-If the outcome is an even number (2, 4, or 6), the player wins $10.
-If the outcome is an odd number (1, 3, or 5), the player loses $5.
Let x be the amount that the player can win or lose in this game. We can represent the possible outcomes of the game and their corresponding probabilities as follows:
-x = $10 with probability 1/2 (rolling 2, 4, or 6)
-x = -$5 with probability 1/2 (rolling 1, 3, or 5)
To compute the mean or expected value of x, we can use the formula:
E(x) = (10)(1/2) + (-5)(1/2) = $2.50
This means that on average, the player can expect to win $2.50 per game if they play this game many times.
To compute the variance and standard deviation of x, we can use the formulas:
Var(x) = E[(x - E(x))^2] = [(10 - 2.50)^2(1/2) + (-5 - 2.50)^2(1/2)] ≈ $26.25
SD(x) = sqrt(Var(x)) ≈ $5.12
These values indicate that there is a relatively large amount of variability in the player's potential winnings and losses in this game, and that the player is more likely to experience losses than gains over the long run. However, because the game is fair (i.e., the expected value is zero), the player is not at an inherent disadvantage and can choose to play or not based on their personal risk tolerance and preferences.