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Binomial Distribution

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data

    http://puu.sh/dOcM [Broken]
    http://puu.sh/dOcZ [Broken]
    2. Relevant equations

    3. The attempt at a solution
    I got Part A.
    For part A, this is what I did:

    I did Egg A: X ~ (6,(1/6)) P(X = 1) and did something similar for Egg B. I then multiplied both to get the answer for Part A.
    http://puu.sh/dOfV [Broken]

    For Part B, I'm a bit confused. I tried doing cases. As in:
    (broken eggs): AA , BB , AB (same as BA)
    Then I did:
    AA: X ~ (6,(1/6)) P(X = 2)
    BB: Y ~ (6,(1/10)) P(X = 2)
    AB: X ~ (6,(1/6)) P(X = 2) * Y ~ (6,(1/10)) P(X = 2)

    My work:
    http://puu.sh/dOh8 [Broken]

    And then, I tried multiplying and adding the three values. But I didn't get the correct answer.

    Could someone help me out please? Thanks
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 18, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If two type A are broken then we must have 0 type B broken in order to have two broken altogether.

  4. Jan 18, 2012 #3
    Wow, thanks. I tried it out. It works :) thanks.
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