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Discrete Math - quick probability questions.

  1. Dec 6, 2015 #1
    For the life of me I am having a hard time understanding how to do problems of this nature. As I understand it, were using the multiplication rule here with a twist.


    a. How many integers from 1 through 100,000 contain the
    digit 6 exactly once?

    5 * 9 * 9 * 9 * 9 = 38805 is what I have. Because the digit 6 can appear in 5 different locations.

    b. How many integers from 1 through 100,000 contain the
    digit 6 at least once?

    c. If an integer is chosen at random from 1 through
    100,000, what is the probability that it contains two or
    more occurrences of the digit 6?

    I am clueless here. Add up all the possibilities of containing two 6's...three 6's... four 6's... five 6's and then divide by the total possibilities.That's my line of thinking... I am just not sure how to find the amount of possibilities for each partition.
     
    Last edited: Dec 6, 2015
  2. jcsd
  3. Dec 6, 2015 #2

    pwsnafu

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    Science Advisor

    How many integers are there that do not contain any digit 6?
     
  4. Dec 6, 2015 #3
    The first thing I always try in probability problems is to invert it. P(X) = 1 - P(not X)
    It's easy to give it a shot, and often the resulting problem is much simpler.

    In this case not X is
    b') How many integers from 1 through 100,000 do not contain digit 6?
    I bet that you can do that.

    Try my special magic rule again.
     
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