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 questions?

  1. Nov 29, 2005 #1
    How to calculate the probability of this questions?

    Supposed there are m bukcets, n balls. use function f(x) to decide which bucket the ball go. Supposed that the probability a ball going to 0-j is 0.9 and going to j -> m-1 is 0.1. What's the probability of for every bucket there is at least one ball?

    I really get stuck.

    Any help appreciated!
     
  2. jcsd
  3. Nov 29, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I do not understand what you mean by "a ball going to 0-j" or "going to j -> m-1 ". I presume that you have the buckets labeled 1 to m by "m-1" you mean the bucket labeled that way- but what is j?
     
  4. Nov 29, 2005 #3
    Yes, the buckets are labeled from 0 to m-1,
    the probaiblity that a ball going to buckets(from 0 to j-1) is 0.9, and going to j-1 to m-1 bucket is 0.1. and inside [0, j-1] and [j-1, m-1] is equally. j is only a number. for example if m=4, j=2 and p(a ball go to bucket 0,1) = 0.9 and p(go to bucket 2,3)=0.1
     
  5. Nov 30, 2005 #4

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    HINT: Calculate the probability that a bucket contains no balls - it is a binomial distribution.

    Also, are you sure the two bucket ranges overlap??
     
  6. Nov 30, 2005 #5
    Thanks a lot. Yes I made a mistake here, the latter should be j -> m-1.
    Your advice is really helpful..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Probability questions?
  1. Probability Question (Replies: 2)

  2. Probability Question (Replies: 12)

  3. Probability Question (Replies: 6)

  4. Probability question (Replies: 10)

  5. Probability Question (Replies: 4)

Loading...