On three rolls of a single​ die, you will lose ​$19 if a 3 turns up at least​ once, and you will win ​$5 otherwise. What is the expected val

Question

On three rolls of a single​ die, you will lose ​$19 if a 3 turns up at least​ once, and you will win ​$5 otherwise. What is the expected value of the​ game?
Let X be the random variable for the amount won on a single play of this game.

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Reagan 1 week 2021-09-15T03:06:46+00:00 2 Answers 0

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    0
    2021-09-15T03:08:04+00:00

    Answer:

    -16 cents

    Step-by-step explanation:

    We are given that on  three rolls of a single​ die, you will lose ​$19 if a 3 turns up at least​ once, and you will win ​$5 otherwise.

    We are to find the expected value of the game.

    P (at least one 5 in three rolls) = 1 – P (no. of 3 in three) = 1-(\frac{3}{6} )^2 = 0.875

    P (other results) = 1 – 0.875 = 0.125

    Random game value = -19, +5

    Probabilities: 0.875, 0.125

    Expected game value (X) = 0.875 × (-19) + 0.125 × (5) = -16 cents

    Therefore, every time you play the game, you can expect to lose 16 cents

    0
    2021-09-15T03:08:07+00:00

    Answer:

    It is expected to lose 5.10 dollars

    Step-by-step explanation:

    The probability of getting a 3 by throwing a die once is 1/6.

    By throwing it 3 times the probability of not getting a 3 is:

    P=(\frac{5}{6}) ^ 3 =0.5787

    Then the probability of obtaining a three at least once in the 3 attempts is:

    P'=(1-0.5787)=0.421

    So if X is the discrete random variable that represents the amount gained in a single move of this game, the expected gain E(X) is:

    E(X)=P'*(X') + P*(X)

    E(X) =0.421'*(-19) + 0.5787*(5)\\\\E(X) =-\$5.10

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