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Probability of finding all 5 errors in a text with 3 people?

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    There are 5 errors, and 3 people are proofreading the text.

    • P(1. person finds an error) = 0,5
    • P(2. person finds an error) = 0,6
    • P(3. person finds an error) = 0,7
    There are 5 errors, and whenever a person comes across one, they have the above probability of seeing it. Depending on the combinations, the probabilities are multiplied together. The question implies that the probabilities are not dependent on each other.

    2. Relevant equations

    P(A and B) = P(A) * P(B)

    3. The attempt at a solution

    Now, my head is telling me, that there are a huge number of ways (combinations) the people could find the errors in. e.g.:

    1. Person 1 finds all errors, the other 2 find none. (0,5^5 * 0,4^5 * 0,3^5)
    2. Person 1 finds 3, person 2 finds 1 and person 3 finds 1. ((0,5^3 * 0,5^2 ) * (0,6 * 0,4^4 ) * (0,7 * 0,3^4 )
    3. Person 1 finds none, person 2 finds 3 and person 3 finds 2. (0,5^5 * (0,6^3 * 0,4^2 ) * (0,7^2 * 0,3^3 ))
    4. All people find all errors. (= 0,5^5 * 0,6^5 * 0,7^5 )
    I know how to calculate combinations, but do I really have to go through all the possibilities and add the probabilities together. Are you even supposed to sum them up?

    There is also the issue of possibly having to consider the order in which the errors are found in, but I very much doubt that is required. That would probably really bloat the answer to unreasonable proportions, and I wouldn't know if that's even possible mathematically.

    For the record, the answer at the back of the book is 0,74. This leads me to think that addition is necessary, since probabilities are always 0 <= P <= 1 and the largest probability of finding an error is less than the answer.
     
  2. jcsd
  3. Feb 10, 2015 #2
    Forget it. Turns out it was a simple solution after all. Got help from another forum, and I can't believe i didn't see the solution right away.

    Just calculate the probability of none of the reviewers finding an error as they come across it, and the reduce that from 1 to get the probability of at least one of them finding it. Then just raise that to the 5th power to extend the probability to all 5 errors.
     
  4. Feb 10, 2015 #3

    BvU

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    It's like the chance of throwing a six with six dice (you'd be surprised at the number of people who go from 1/6, 2/6, 3/6 to 6/6 if you bring it to them gradually).
     
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