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Dependent theorem and bayes theorem.

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    One box of eggs has 12 eggs. The probability of all eggs being good is 70%. The probability of one being bad is 20%. The probability of two being bad is 10%. Given that 2 randomly selected eggs are good, what is the probability that all the eggs will be good?

    2. Relevant equations

    3. The attempt at a solution

    According to bayes theorem,

    P(all good | random 2 are good) = ( P(random 2 are good | all good) * P(all good) ) / P(random two are good)

    Now, P(random 2 are good | all good) = P(random 2 are good <and> all good) / P(all good)

    => " P(random 2 are good <and> all good) " are dependent events. Can't seem to figure out how to get around this.
  2. jcsd
  3. Sep 22, 2012 #2

    Ray Vickson

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    For b = 0,1,2, what are P(random 2 are good|12-b good and b bad)? So, what is P(random 2 are good)?

  4. Sep 23, 2012 #3
    I am currently working on a very very similar question on my own. But, the problem I have is finding the P(Random 2 are good).

    Needed Information
    P(all are good) = 0.7
    P(Random 2 are good | all are good) = 1 (since given all are good, any random ones will be guranteed to be good)
    P(Random 2 are good) This is where I struggle, how to find this?
  5. Sep 23, 2012 #4

    Ray Vickson

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    Read the question immediately above this one. Have you done the two other computations? Until you do that, you will be unable to get anywhere. I will just give one more hint: what is P(both good|1 bad, 19 good in carton)? Well, you are GIVEN that the carton has 19 good eggs in it, and one bad egg. You draw 2 eggs at random, and want the probability that both are good. Surely you must have seen or done such problems before, but if not: (i) what is the probability the first drawn egg is good? (ii) given that the first egg is good, we now are left with 19 eggs (18 good, one bad), so now what is the probability of drawing another good egg?

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