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Math Help - Distribute, Expect!

  1. #1
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    Distribute, Expect!

    Got a simple question i am trying to solve guys, could definitely do with some help. I am very glad my question will come easy to you guys.

    Two numbers are chosen at random from {1,2,3} with repetitions allowed. Let the random variable X denote the sum of the 1st number and 2 times the 2nd number.

    (i) Find the Distribution of X.
    (ii) Find the Expectation E(X) of X.

    Thank you, bad math student trying to catch up
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  2. #2
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    I cant see a clever method to do this, so i'd do it the old fashioned way.

    there are only 18 possible outcomes, each with probability 1/18.

    Code:
    Chosen    |   X   | Prob
    1,1,1     |   2   | 1/18
    1,1,2     |   4   | 1/18
    1,1,3     |   6   | 1/18
    2,1,1     |   2   | 1/18
    ..... etc
    As an example, from the incomplete table above, we have:

    P(X=2) = 2/18
    (it appears twice, each with probability 1/18)

    When you do the full table it will appear more times.

    Use the same method to work out the chance of X taking all possible values.
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  3. #3
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    Quote Originally Posted by SpringFan25 View Post
    ...we have:

    P(X=2) = 2/18
    (it appears twice, each with probability 1/18)
    "Let the random variable X denote the sum of the 1st number and 2 times the 2nd number." I would guess that the first value of X would be '3' since the first number is one, added with 2 times 1 with a result of two giving 3 as an answer.

    I sound like a smartass, but i am getting nowhere >_<
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  4. #4
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    Quote Originally Posted by Aeonitis View Post
    Got a simple question i am trying to solve guys, could definitely do with some help. I am very glad my question will come easy to you guys.

    Two numbers are chosen at random from {1,2,3} with repetitions allowed. Let the random variable X denote the sum of the 1st number and 2 times the 2nd number.

    (i) Find the Distribution of X.
    (ii) Find the Expectation E(X) of X.

    Thank you, bad math student trying to catch up
    (1,1)...1+2(1)=3
    (2,1)...2+2(1)=4
    (3,1)...3+2(1)=5
    (1,2)...1+2(2)=5
    (2,2)...2+2(2)=6
    (3,2)...3+2(2)=7
    (1,3)...1+2(3)=7
    (2,3)...2+2(3)=8
    (3,3)...3+2(3)=9

    Finally E(X) is the mean or average of X (the sums)
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  5. #5
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    Awesome Archie, and that is the full answer, THAT SIMPLY?! oh, just making sure, i mean the right answer would be that the expectation of X i.e. E(X) is 54/9 giving a result of 6?!
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  6. #6
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    Quote Originally Posted by Aeonitis View Post
    Awesome Archie, and that is the full answer, THAT SIMPLY?! oh, just making sure, i mean the right answer would be that the expectation of X i.e. E(X) is 54/9 giving a result of 6?!
    Yes, that's it..
    As the saying goes..."it's easy when you know how"...

    Or when you understand the lexicon.
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