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    please help (( I know how to calculate the expected mean but I don't know how to work out the probability distribution for both which I need.

    I have to leave the house in like 40 minutes and this is due in my first lesson (((

    What I get for leaving homework until last minute :grumble:
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    (Original post by bluemadhatter)
    please help (( I know how to calculate the expected mean but I don't know how to work out the probability distribution for both which I need.

    I have to leave the house in like 40 minutes and this is due in my first lesson (((

    What I get for leaving homework until last minute :grumble:
    Well, you know that dices can give scores of 1, 2, 3, 4, 5 or 6.

    So if you throw two dices, the product of their two scores could be anywhere from 1*1, 1*2, 2*5, 6*6, etc... so you'll need to list each one of those down and calculate the probability of it.

    1*1 is the score of 1 on the first die and a score of 1 on the second die, so that's a probability of 1/6 * 1/6 = 1/36
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    (Original post by Zacken)
    Well, you know that dices can give scores of 1, 2, 3, 4, 5 or 6.

    So if you throw two dices, the product of their two scores could be anywhere from 1*1, 1*2, 2*5, 6*6, etc... so you'll need to list each one of those down and calculate the probability of it.

    1*1 is the score of 1 on the first die and a score of 1 on the second die, so that's a probability of 1/6 * 1/6 = 1/36
    I don't understand :sad:

    For 8a) Its asking the probability distribution of Y, the score on a player's turn
    and its included all the heads and tails bit as well

    and for 9a) It asks for the probability distribution of X (the product of the scores on the top faces)

    hmm so is the probability of each score not 1/6 so the probability of each product is 1/6*1/6 = 1/36 does that mean its a uniform distribution?? hmm maybe I DID get this one???
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    (Original post by bluemadhatter)
    I don't understand :sad:

    For 8a) Its asking the probability distribution of Y, the score on a player's turn
    and its included all the heads and tails bit as well

    and for 9a) It asks for the probability distribution of X (the product of the scores on the top faces)

    hmm so is the probability of each score not 1/6 so the probability of each product is 1/6*1/6 = 1/36 does that mean its a uniform distribution?? hmm maybe I DID get this one???
    Oh, sorry. I was answering only question 9. It is indeed a uniform distribution for that one. Each product has a probability of 1/36.
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    (Original post by Zacken)
    Oh, sorry. I was answering only question 9. It is indeed a uniform distribution for that one. Each product has a probability of 1/36.
    Okay thank you, question 9 done Could you help me with 8a please??
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    (Original post by bluemadhatter)
    Okay thank you, question 9 done Could you help me with 8a please??
    Okay, so we know that the possibilites are 1, 2, ..., 6 just like in the table drawn for you. It'll be the same table, just with different probabilities for each X.

    So, for example getting a 1: The probability of this is given by getting a heads, rolling one die and getting a 1 (1/2 * 1/6) added to the probability of getting a tails, rolling two dices and getting a (as the highest score) 1. (1/2 * 1/36) where the 1/36 comes from the table.

    Getting a 2: The probability of this is given by getting a heads, rolling one die and getting a 2 (1/2 * 1/6) added to the probability of getting a tails, rolling two dices and the highest score being a 2 (1/2 * 3/36)

    Getting a 3: etc...
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    (Original post by Zacken)
    Okay, so we know that the possibilites are 1, 2, ..., 6 just like in the table drawn for you. It'll be the same table, just with different probabilities for each X.

    So, for example getting a 1: The probability of this is given by getting a heads, rolling one die and getting a 1 (1/2 * 1/6) added to the probability of getting a tails, rolling two dices and getting a (as the highest score) 1. (1/2 * 1/36) where the 1/36 comes from the table.

    Getting a 2: The probability of this is given by getting a heads, rolling one die and getting a 2 (1/2 * 1/6) added to the probability of getting a tails, rolling two dices and the highest score being a 2 (1/2 * 3/36)

    Getting a 3: etc...
    where did this value come from? Is it not 2/36 or 1/36. How did you work out the probability of getting the highest score as 2?

    Sorry >.<
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    (Original post by bluemadhatter)
    where did this value come from? Is it not 2/36 or 1/36. How did you work out the probability of getting the highest score as 2?

    Sorry >.<
    In the table given to you, it says "when two fair dice are shown the probability of the highest score being X is shown below". You want P(highest score is 2) in that case, which si the number number 2 in your table.
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    (Original post by Zacken)
    In the table given to you, it says "when two fair dice are shown the probability of the highest score being X is shown below". You want P(highest score is 2) in that case, which si the number number 2 in your table.
    OOHHHHHHHH OKAY I get it now!!! Thank you so much :hugs:
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    (Original post by bluemadhatter)
    OOHHHHHHHH OKAY I get it now!!! Thank you so much :hugs:
    Glad you got it! Buuut, I think what I said about Q9 is wrong. Groggy mornings and all.

    It can't be uniform because some of the numbers can't be reached, i.e: no product will be "7", so they probably want you to find all the possible products and find each probability individually. Sorry.
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    (Original post by Zacken)
    Glad you got it! Buuut, I think what I said about Q9 is wrong. Groggy mornings and all.

    It can't be uniform because some of the numbers can't be reached, i.e: no product will be "7", so they probably want you to find all the possible products and find each probability individually. Sorry.
    Its fine at least I've "attempted" it :rofl: Now I must run out the house O.O

    Thanks again
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    (Original post by bluemadhatter)
    Its fine at least I've "attempted" it :rofl: Now I must run out the house O.O

    Thanks again
    Glad I "helped". :rofl:
 
 
 
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