Question 604734: For a certain game of chance, a player loses $3 with a probability of 0.3, breaks even with probability 0.3, gains $1 with probability 0.2, gains $2 with probability 0.1, and gains $3 with probability 0.1.
) A player plays the game of chance. What is the probability that a player will win some money? Show work.
) If the player plays the game many times, what is the player’s expectation?
That is, what is the expected value (mean) of the probability distribution? Show work. (You are welcome to use the extra column and/or row in the table to make it easier to show the computation.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! For a certain game of chance, a player loses $3 with a probability of 0.3, breaks even with probability 0.3, gains $1 with probability 0.2, gains $2 with probability 0.1, and gains $3 with probability 0.1.
) A player plays the game of chance. What is the probability that a player will win some money? Show work.
P(winning) = 0.2+0.1 = 0.3
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) If the player plays the game many times, what is the player’s expectation?
E(x) = -3*0.3 + 0*3 +1*0.2 + 2*0.1 + 3*0.1
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= -0.9 + 0 + 0.2 + 0.2 + 0.3
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= -0.9 + 0.7
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= -0.2
Means the player can expect to lose 20 cents each time he plays the game.
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cheers,
Stan H.
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