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Math Help - Bernoulli trial with conditional probability

  1. #1
    jut
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    Bernoulli trial with conditional probability

    In a communication system the probability, denoted Pr(), is

    Pr(0 being transmitted) = 0.4
    Pr(1 being transmitted) = 0.6

    Pr(receive 0 | transmit 0) = 0.92
    Pr(receive 1 | transmit 1) = 0.95
    Notation: Given that a 0 was transmitted, prob of receiving a 0 is 0.92

    What's the prob of receiving 6 successive symbols without error?

    My first task, I suppose, is to find out the prob of receiving with no error. Would that prob equal

    Pr(receive 0 | transmit 0) + Pr(receive 1 | transmit 1) - Pr(receive 0 | transmit 0) * Pr(receive 1 | transmit 1) = 0.996

    since they're mutually exclusive? That seems wrong because it's > than the individual prob of 0.92 or 0.95. I also need to factor in the prob of each symbol being transmitted also.
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  2. #2
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    Quote Originally Posted by jut View Post
    In a communication system the probability, denoted Pr(), is

    Pr(0 being transmitted) = 0.4
    Pr(1 being transmitted) = 0.6

    Pr(receive 0 | transmit 0) = 0.92
    Pr(receive 1 | transmit 1) = 0.95
    Notation: Given that a 0 was transmitted, prob of receiving a 0 is 0.92

    What's the prob of receiving 6 successive symbols without error?

    My first task, I suppose, is to find out the prob of receiving with no error. Would that prob equal

    Pr(receive 0 | transmit 0) + Pr(receive 1 | transmit 1) - Pr(receive 0 | transmit 0) * Pr(receive 1 | transmit 1) = 0.996

    since they're mutually exclusive? That seems wrong because it's > than the individual prob of 0.92 or 0.95. I also need to factor in the prob of each symbol being transmitted also.
    The probability of receiving one symbol without error is

    Pr(receive 0 | transmit 0).Pr(transmit 0) + Pr(receive 1 | transmit 1).Pr(transmit 1) = (0.92)(0.4) + (0.95)(0.6) = 0.938.

    Receiving each signal is an independent event therefore Pr(receiving 6 successive symbols without error) = (0.938)^6 = ....
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  3. #3
    jut
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    Quote Originally Posted by mr fantastic View Post
    The probability of receiving one symbol without error is

    Pr(receive 0 | transmit 0).Pr(transmit 0) + Pr(receive 1 | transmit 1).Pr(transmit 1) = (0.92)(0.4) + (0.95)(0.6) = 0.938.
    I'm having a hard time with that. For a given signal, if i receive a 1, then that excludes getting a zero. If I receive a 0, that excludes getting a 1. So wouldn't those events be mutually exclusive? So wouldn't the probability of receiving one symbol without error be:

    Pr(Rx 0 | Tx 0).Pr(Tx 0) + Pr(Rx 1 | Tx 1).Pr(Tx 1) - [Pr(Rx 0 | Tx 0).Pr(Tx 0) * Pr(Rx 1 | Tx 1).Pr(Tx 1)]
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