1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Probability of getting 3 on loaded die
  1. Probability in a die (Replies: 0)