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Probability of getting 3 on loaded die

  1. Oct 30, 2005 #1
    A six-sided die is loaded in such a way that an odd number is twice as likely to occur as an even number. We throw the die until a 3 is observed. What is the probability that the number of throws required is (strictly) more than 5 and (strictly) less than 10?

    Answer: 0.1805.

    My working.

    P(odd) = 2P(even)
    P(odd) + P(even) = 1
    P(odd) = 2/9.

    X = "number of throws before success"

    X is geometric with p = 2/9.


    [itex]P( 6\leq X \leq 9 ) = \sum_{i=6}^9 (2/9)(1-2/9)^i = 0.1403642 [/itex].

    Any help would be appreciated.


  2. jcsd
  3. Oct 30, 2005 #2
    I finally figured it out. For anyone interested, the solution involves a change of variables from [itex]X[/itex] (the number of failures before success) to [itex]Y = X + 1[/itex] (the number of attempts required for success).
    So [itex]P( 6\leq X \leq 9 ) = \sum_{i=6}^9 (2/9)(1-2/9)^{i-1} = 0.1805 [/itex].
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