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Math Help - [SOLVED] Independent Events

  1. #1
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    [SOLVED] Independent Events

    hai...stuck ..it simple but i still stuck. hope some one give me guide line how to solve

    Question:
    A coin is tossed three times. A is the event ' at least two tails' , B isthe event 'three heads or three tails, C is the event 'at least one tail. Which of the following are independent?

    (a) A and B
    (b) A and C
    (c) B and C

    solution:
    (a) yes
    (b) no
    (c) no

    so how to prove the ans
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  2. #2
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    Hi

    Compute p(A), p(B), p(C), p(A and B) ...
    and show that p(A and B)=p(A).p(B)
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  3. #3
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    Quote Originally Posted by nikk View Post
    Question:
    A coin is tossed three times. A is the event ' at least two tails' , B isthe event 'three heads or three tails, C is the event 'at least one tail. Which of the following are independent?
    (a) A and B
    (b) A and C
    (c) B and C
    solution:
    (a) yes (b) no (c) no
    You need to show that P(A \cap B)=P(A)P(B)\;\&\;P(A \cap C)<br />
\not{=}P(A)P(C)
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  4. #4
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    My outcomes as below:
    HHH,HHT,HTH,HTT,THH,THT,TTH,TTT

    Question (a)
    i can calculate P(A) = P(HTT) = 1/2 * 1/2 * 1/2 = 1/8
    = P(THT) = 1/2 * 1/2 * 1/2 = 1/8
    = P(TTH) = 1/2 * 1/2 * 1/2 = 1/8
    = P(TTT) = 1/2 * 1/2 * 1/2 = 1/8

    There for P(A) = P(HTT) + P(THT) + P(TTH) + P(TTT)
    = 1/8 + 1/8 + 1/8 + 1/8
    = 4/8
    = 1/2
    is it correct P(A) = 1/2????


    P(B) = P(HHH) + P(TTT)
    = 1/8 + 1/8
    = 1/4

    p(A and B)= P (TTT) + P (TTT)
    = 1/8 + 1/8
    = 1/4

    So, p(A and B) = P(A) * P(B)
    1/4 = 1/2 * 1/4
    = 1/8

    so, not independent but the ans is independent
    Last edited by nikk; February 9th 2009 at 06:04 AM. Reason: P(B) and final ans
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  5. #5
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    logic

    is it my equation of P(A) = P(HTT) + P(THT) + P(TTH) + P(TTT)correct
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  6. #6
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    Quote Originally Posted by nikk View Post
    is it my equation of P(A) = P(HTT) + P(THT) + P(TTH) + P(TTT)correct
    Yes it is correct.
    But you could make a table, like a truth table.
    \begin{array}{ccccl} H &\vline &  H &\vline &  H  \\   H &\vline &  H &\vline &  T  \\   H &\vline &  T &\vline &  H  \\   H &\vline &  T &\vline &  T{\color{red}\leftarrow}  \\   T &\vline &  H &\vline &  H  \\   T &\vline &  H &\vline &  T{\color{red}\leftarrow}  \\   T &\vline &  T &\vline &  H{\color{red}\leftarrow}  \\   T &\vline &  T &\vline &  T {\color{red}\leftarrow} \\ \end{array}

    Note that there are 2^3 =8 rows for the three coins.
    Now the counting is easy, see the arrows. We get \frac{4}{8} = \frac{1}{2}
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