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Math Help - Some revision needs checking - Simple questions.

  1. #1
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    Some revision needs checking - Simple questions.

    Hi,

    I'm doing some revision on an upcoming test and would like for someone to runover my work to see if its correct. Thanks..



    Is it okay to submit my work like this..I'm not sure of another way?

    Thanks..
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  2. #2
    Super Member
    Joined
    May 2009
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    In the 2nd problem there are 32 (2^5) elements in the outcome set, not 6.

    Let's look at a simpler variant of this problem. If you were tossing a coin 3 times, there would be 8 (2^3) elements in the outcome set:
    O = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
    Then
    P(at most half the coin tosses are heads)
    = P(exactly 0 heads) + P(exactly 1 head)
    = 1/8 + 3/8
    = 1/2

    i) For tossing a coin 5 times, the outcome set starts like this:
    O = {HHHHH, HHHHT, HHHTH, HHHTT...}
    Then
    P(at most half the coin tosses are heads)
    = P(exactly 0 heads) + P(exactly 1 head) + P(exactly 2 heads)
    = ?/32 + ?/32 + ?/32
    ...

    Can you take it from here?

    ii) This is the easier one. The answer would be 1/32.


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  3. #3
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    Dec 2007
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    Thanks yeongil..

    I'm going to organise some tuition, god knows I need it.

    Just a quick question... Do I need to write out the full outcome set to be able too fill in your question marks?

    Thanks again.
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  4. #4
    Newbie
    Joined
    Dec 2007
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    Alas, here I am.back again.

    So what I did was I drew out a complete truth table for the outcome set..

    P(0H) + P(1H) + P(2H)

    1/32 + 3/32 + 9/32

    => 13/32

    I did this a shamefully long way.. I had to draw out the complete table and manually lookup the number of times each event occured. Is there a quicker way of doing this?

    Thanks for looking over these for me as I have a repeat exam in a few days time and I think i'll struggle to get 40% (Shame on me I know lol)

    Thanks..
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