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Unique Dice Throwing Probability Question

  1. Oct 8, 2014 #1
    G'day,

    I've recently been given a probability question to solve and I'm not 100% on my approach towards it.
    It goes as follows;

    A dice is found to be weighted so that the chance of throwing a 6 is four times the chance of throwing a 1. Chances of throwing numbers other than 6 are equally likely.

    a) What is the probability of throwing a 6?
    b) What is the probability of throwing a 3?
    c) What is the probability of throwing at least a 3?


    For my approach I'm basically assuming the chance of throwing anything except a six is 1/6, which means the chance of throwing a 6 if 4/6.
    So that answers part a) and b), although I feel that the answer is to simple.
    And then to answer part c) I just multiply the chances of throwing a 3, 4, 5, 6 independently, and then sum those probabilities. .

    Any comments on this approach or help with the question would be greatly appreciated.
     
    Last edited: Oct 8, 2014
  2. jcsd
  3. Oct 8, 2014 #2
    What should the sum of the probabilities of all of the possible outcomes be? Is that true in your model?
     
  4. Oct 9, 2014 #3
    Well initially I thought the probability would be (1/6)*5 + (4/6) => 9/6

    The 1/6 being the chance of rolling a 1-5
    The 4/6 being the chance of rolling a 6 (It being 4 times more likely)

    But then I came up with a more logical answer;

    Probability of rolling a 1-5
    x = P(D1-5)

    Probability of rolling a 6
    4x = P(D6)

    BUT => P(D1-5) + P(D6) =1

    Therefore;
    x + 4x = 1
    5x = 1
    x = 1/5

    So we end up with

    P(D1-5) = 1/5
    P(D6) = 4/5

    So my new answers are
    part a) (4/5)
    part b) (1/5)

    For part c) I've come up with

    P(D1-2) = 1/5*2
    P(D3-5) = 1/5*3
    P(D6) = 4/5

    P(at least a 3) = P(D3-5) + P(D6) - P(D1-2)
    P(D>=3) = 3/5 + 4/5 - 2/5 = 1

    I am on the right track with this?
     
    Last edited: Oct 9, 2014
  5. Oct 9, 2014 #4

    statdad

    User Avatar
    Homework Helper

    "A dice is found to be weighted so that the chance of throwing a 6 is four times the chance of throwing a 1. Chances of throwing numbers other than 6 are equally likely."

    Your question isn't clearly written. There are two ways to interpret these parts.
    a) If p1 = a, then p6 = 4a. If the chances of throwing numbers other than 6 are equally likely is taken directly, this means that also p2 = p3 = p4 = p5 = a as well. If this is true, then it is very easy to find a.

    b) If p1 = a, then p6 = 4a. If the second statement applies only the the numbers 2, 3, 4, 5, then all you have from that is that p2 = p3 = p4 = p5 = b, and you don't have enough information to find both a and b.

    I will guess you mean the first situation, but you need to be clear about it.
     
  6. Oct 9, 2014 #5
    What should ##P(D_6)## be in terms of ##P(D_1)## ?

    What should ##P(D_2)## be in terms of ##P(D_1)## ?

    What should ##P(D_1)+P(D_2)+P(D_3)+P(D_4)+P(D_5)+P(D_6)## be?
     
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