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Math Help - help with probability!! #2

  1. #1
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    Exclamation help with probability!! #2

    I need help with the following questions, i attempted to solve them, and i also put my answers here too.
    When you are answering them, please show all your steps as i am kinda lost in this field.

    1. suppose S = {1,2,3} and P({1,2}) = 1/3 and P({2,3}) = 2/3
    compute P({1}), P({2}), P({3})
    my answer
    P({1}) + P({2}) + P({3}) = 1 <--- 1 being 100% and 0 being 0%
    P({1}) + P({2}) = 1/3
    P({2}) + P({3}) = 2/3
    a) P({1}) + (1/3 - P({1})) + P({3}) = 1
    P({3}) = 2/3
    b) P({1}) + (2/3 - P({3})) + P({3}) = 1
    P({1}) = 1/3
    so then 1/3 + P({2}) + 2/3 = 1
    so P({2}) = 0
    therefore: P({1}) = 1/3, P({2}) = 0, P({2}) = 2/3

    2. Suppose A1 watches the six o'clock news a 2/3 of the time and watches eleven o'clock news 1/2 of the time and watches both news 1/3 of the time. For a randomely selected day what is the probability that A1 watches only the six o'clock new?s, and whats the probability that A1 watches neither news?
    my answer
    chance for 6oclock = 2/3
    chance for 11oclock news = 1/2 => complementary is 1/2
    for only six o'clock news
    has to watch the 6oclock news which is 2/3, and not watch the 11oclock news which is 1/2
    2/3*1/2 = 1/3
    so only watching 6 o'clock news is 1/3 chance...
    for watching neither
    avoid watching 6o'clock news which is 1/3 (complementary) and avoid watching eleven o'clock news 1/2 (complementary)
    so its
    1/3*1/2 = 1/6
    so watching neither is 1/6 chance...
    Last edited by mr fantastic; September 15th 2010 at 08:44 PM. Reason: Moved.
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