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Math Help - Probility questions:

  1. #1
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    galway
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    Probility questions:

    doing some probability questions for a assignment, I think i have them right but maths isnt my strong suit could anyone look at these two questions comment if these are correct

    A computer randomly chooses 4 different letters from the 26 letters of the alphabet.
    Calculate the probability that the chosen letters spell the word ZONE

    solution:
    26*26*26*26 =456976
    p(zone) is 1 over 456976

    ---------------------------------------

    Suppose A and B are events with P(A) = 0.4 and P(B) = 0.4.
    Find the probability that
    (i) A does not occur
    (ii) B does not occur
    (iii) A or B occurs
    (iv) Neither A nor B occurs

    Solution:
    (i) P(not A)= 1-0.4 = 0.6 = 3 over 5
    (ii) P(not B) = 1 - 0.4 = 0.6 = 3 over 5
    (iii)P(AorB) = 0.4 +0.4 =0.8 = 4 over 5
    (iv) P(not A or B) = 1 - 0.8 =0.2 = 1 over 5 (*sorry i dont know hoe to edit this to show fractions)

    thanks in advance .
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  2. #2
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    Re: Probility questions:

    Quote Originally Posted by ronanbrowne88 View Post
    doing some probability questions for a assignment, I think i have them right but maths isnt my strong suit could anyone look at these two questions comment if these are correct.

    A computer randomly chooses 4 different letters from the 26 letters of the alphabet.
    Calculate the probability that the chosen letters spell the word ZONE
    solution:
    [/B]26*26*26*26 =456976
    p(zone) is 1 over 456976
    Read the question carefully. It says 4 different letters. So it it would be 26\cdot 25\cdot 24 \cdot 23.


    Quote Originally Posted by ronanbrowne88 View Post
    Suppose A and B are events with P(A) = 0.4 and P(B) = 0.4.
    Find the probability that
    (i) A does not occur
    (ii) B does not occur
    (iii) A or B occurs
    (iv) Neither A nor B occurs

    Solution:
    (i) P(not A)= 1-0.4 = 0.6 = 3 over 5
    (ii) P(not B) = 1 - 0.4 = 0.6 = 3 over 5
    (iii)P(AorB) = 0.4 +0.4 =0.8 = 4 over 5
    (iv) P(not A or B) = 1 - 0.8 =0.2 = 1 over 5
    You have answered parts (i) & (ii) correctly.

    However the is no way possible to answer parts (iii) & (iv).
    There is not enough information given.
    We need to know if A~\&~B are independent or else the value of \mathcal{P}(A\cap B).
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  3. #3
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    Re: Probility questions:

    i think they are independent, if that is the case is the above correct
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  4. #4
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    Re: Probility questions:

    Quote Originally Posted by ronanbrowne88 View Post
    i think they are independent, if that is the case is the above correct
    No. Part (iii) is \mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(B)-\mathcal{P}(A\cap B).
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