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Math Help - [SOLVED] help please

  1. #1
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    Exclamation [SOLVED] help please

    There are two stations from A to B and two stations from B to C. Suppose that each station is closed with probability p and that the state of each station is independent of the others. What condition on p ensures that the probability that I cannot travel from A to C on train is no more than 1/2?


    Can some please guide me on doing this probability problem.
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  2. #2
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    The probability that any station is open is (1-p).

    The probability that all stations are open is (1-p)^4.

    You want the likelihood that you can travel to be

    (1-p)^4 \geq \frac{1}{2}
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  3. #3
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    [quote=Henderson;209102]The probability that any station is open is (1-p).

    The probability that all stations are open is (1-p)^4.

    You want the likelihood that you can travel to be

    (1-p)^4 \geq \frac{1}{2}[/quote}

    So what will the probability that i can not travel be? I don't understand?
    Last edited by maths2008; October 25th 2008 at 10:21 AM.
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  4. #4
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    Hello, maths2008!

    There are two stations from A to B and two stations from B to C.
    . .
    Are there three stations or four?
    Suppose that each station is closed with probability p
    and that the state of each station is independent of the others.
    What condition on p ensures that the probability that
    I cannot travel from A to C on train is no more than 1/2?
    I will assume there are three stations.
    Code:
          * - - - - - * - - - - - - *
          A           B             C

    We cannot make the trip if one or more stations are closed.
    The only way to make the trip is when all three stations are open.

    Let q \:=\:1-p be the probability that a station is open.

    The probability that all three stations are open is: q^3
    . . and we want this probability to be greater than one-half.

    So we have: . q^3 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[3]{2}}

    Hence: . p \;\geq\;1-q \;=\;1-\frac{1}{\sqrt[3]{2}}  \;=\;0.206299474

    Therefore: . p \:\geq \:0.2063
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, maths2008!

    I will assume there are three stations.
    Code:
          * - - - - - * - - - - - - *
          A           B             C
    We cannot make the trip if one or more stations are closed.
    The only way to make the trip is when all three stations are open.

    Let q \:=\:1-p be the probability that a station is open.

    The probability that all three stations are open is: q^3
    . . and we want this probability to be greater than one-half.

    So we have: . q^3 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[3]{2}}

    Hence: . p \;\geq\;1-q \;=\;1-\frac{1}{\sqrt[3]{2}} \;=\;0.206299474

    Therefore: . p \:\geq \:0.2063
    Thanks for the help but actually there are 4 stations

    Shouldn't i caluclate the probabilty that i can not travel to A to C as the question suggests that What condition on P ensures that the probability that
    I cannot travel from a to c is no more than 1/2?
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  6. #6
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    Hello again, maths2008!

    So there are four stations.
    Code:
          * - - - - - * - * - - - - - - *
          1           2   3             4

    We cannot make the trip if one or more stations are closed.
    The only way to make the trip is when all four stations are open.

    Let q \:=\:1-p be the probability that a station is open.

    The probability that all four stations are open is: q^4
    . . and we want this probability to be greater than one-half.

    So we have: . q^4 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[4]{2}}

    Hence: . p \;\leq\;1-q \;=\;1-\frac{1}{\sqrt[4]{2}}  \;=\;0.159103585

    Therefore: . p \:\leq \:0.1591


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    You can solve the problem "head on" if you dare.

    But there are fifteen cases you must consider . . .

    . . C \:=\:\text{Closed, }\;O \:=\:\text{O{p}en}

    . . \begin{array}{|c|c|c|c| } 1&2&3&4 \\ \hline \hline C&C&C&C \\ C&C&C&O \\ C&C&O&C \\ C&C&O&O \\ C&O&C&C \\ C&O&C&O \\ C&O&O&C \\ C&O&O&O \\ \end{array}. . . \begin{array}{|c|c|c|c|} 1&2&3&4 \\ \hline O&C&C&C \\ O&C&C&O \\ O&C&O&C \\ O&C&O&O \\ O&O&C&C \\ O&O&C&O \\ O&O&O&C \end{array}

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  7. #7
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    Smile

    Quote Originally Posted by Soroban View Post
    Hello again, maths2008!

    So there are four stations.
    Code:
          * - - - - - * - * - - - - - - *
          1           2   3             4
    We cannot make the trip if one or more stations are closed.
    The only way to make the trip is when all four stations are open.

    Let q \:=\:1-p be the probability that a station is open.

    The probability that all four stations are open is: q^4
    . . and we want this probability to be greater than one-half.

    So we have: . q^4 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[4]{2}}

    Hence: . p \;\leq\;1-q \;=\;1-\frac{1}{\sqrt[4]{2}} \;=\;0.159103585

    Therefore: . p \:\leq \:0.1591


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    You can solve the problem "head on" if you dare.

    But there are fifteen cases you must consider . . .

    . . C \:=\:\text{Closed, }\;O \:=\:\text{O{p}en}

    . . \begin{array}{|c|c|c|c| } 1&2&3&4 \\ \hline \hline C&C&C&C \\ C&C&C&O \\ C&C&O&C \\ C&C&O&O \\ C&O&C&C \\ C&O&C&O \\ C&O&O&C \\ C&O&O&O \\ \end{array}. . . \begin{array}{|c|c|c|c|} 1&2&3&4 \\ \hline O&C&C&C \\ O&C&C&O \\ O&C&O&C \\ O&C&O&O \\ O&O&C&C \\ O&O&C&O \\ O&O&O&C \end{array}

    Hi Soroban

    Thanks for the help.
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