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Math Help - lottery game question

  1. #1
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    lottery game question

    Question. Consider the following lottery game involving two urns:
    - Urn A: five balls, labelled 1 to 5
    - Urn B: five balles, four green and one red.

    Tickets cost GBP1 and players select to numbers between 1 and 5 (repetition not permitted). The lottery draw consists of two balls taken without replacement from Urn A, and one ball taken from Urn B. For a given ticket, there are three possible outcomes:

    - numbers on ticket do not match draw from Urn A ->win GBP 0
    - numbers on ticket match draw from Urn A and a green ball is drawn from Urn B ->win GBP1
    - numbers on ticket match draw from Urn A and a red ball is drawn from Urn B -> win GBP (z+1).

    (a) evaluate probability of each of these three outcomes.

    (b) define random variable R as the return from playing the game with a single ticket; so R is our winnings minus the GBP 1 cost of the ticket. Write down the probability mass function of R and evaluate the expected return E(R). What value must z take for the expected return to be zero?

    Answer.
    (it is a simple question but I almost always make mistakes in questions involving evaluating probabilities... could you pls check my answer).

    (a) probability of each outcome

    Urn A: total number of ways to pick 2 numbers out of 5 is 5(Choose)2 \frac{5!}{2!(5-2)!}=10. Only one combination is a 'match' therefore

    P(match 2 numbers in Urn A)=1/10.

    P(not match two numbers in A)=1-1/10-9/10= P(win GBP 0) (return is 0-1=-1)

    P(match urn A AND Green ball)=1/10*4/5 (independent events)=4/50. (Return is 1-1=0)

    P(match urn A AND red ball)=1/10*1/5=1/50 (Return is z+1-1=z)

    (b) Return mass function shows probability of each possible random value ie

    P(R=-1)=9/10
    P(R=0)=4/50
    P(R=z)=1/50

    Through applying these probabilities to each return value and finding the sum of all products, we get E(R)=0 when z=45.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    All seems good to me
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  3. #3
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    Oh really? I am pleasantly surprised ))) Thanks!
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