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Probablility question

  • 08-05-2013 1:34pm
    #1
    Daniel S
    Registered Users, Registered Users 2 Posts: 3,533 ✭✭✭


    The chance of an own-goal in a match is 1 in 80.
    In 80 matches, what is the probability of:
    (i) No own-goal.
    (ii) At least one own-goal.
    (iii) What is the Expected value for the number of own goals?

    The last two parts are fine, but I can't get my head around the first part. I know it a binomial trial, but I'm just getting 1 as my answer, which can't be right.

    For the last two parts, it's just 1 minus the answer from the first part and for part (iii) n x p = (1/80) x 80 = 1


Comments

  • #2
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    Assuming that all these are independent, then the probabilities of getting no own goal in each match (79/80) can be multiplied repeatedly to get the probability of getting no own goal in any of the 80 matches. This gives: (79/80)^80. (about 0.366).


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