If a 10 symbol sequence is sent through the channel,what is the probability that up to 3 symbols are in error out of the 10transmissions. Sh

Question

If a 10 symbol sequence is sent through the channel,what is the probability that up to 3 symbols are in error out of the 10transmissions. Show how to make a calculation to obtain a final value.

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Bella 2 hours 2021-10-13T02:49:43+00:00 1 Answer 0

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    2021-10-13T02:51:01+00:00

    Answer: 0.171887

    Step-by-step explanation:

    Given that S0 and S1 are binary symbol of equal probabilty;

    P(S0) = P(S1) = 0.5

    This probability is a Binomial random variable of sequence Sn, where Sn counting the number of success in a repeated trials.

    P(Sn =X) = nCx p^x (1-p)^(n-1)

    Pr(at most 3) = P(0<= x <=3) = P(X=0) + 0) + P(X=1) + P(X=2) + P(X=3)

    Since there are only 2 values that occur in sequence 0 and 1 ( or S0 and S1).Let the distribution be given by the sequence (0111111111),(1011111111),(11011111111),…(1111111110) for Sn= 1 is the sequence for 1 error.

    10C0, 10C1, 10C2, and 10C3 is the number of sequences in value for X= 0, 1, 2, 3 having value 0 and others are 1. Let the success be p(S0)=0.5 and p(S1)= 0.5

    P(0<= X <=3) = 10C0 × (0.5)^0 × (0.5)^10 + 10C1× (0.5)¹ × (0.5)^9 + 10C2 × (0.5)² × (0.5)^8 + 10C3(0.5)³(0.5)^7

    = 1 × (0.5)^10 + 10 × (0.5)^10 + 45 × (0.5)^10 + 120 × (0.5)^10

    =0.000977 + 0.00977 + 0.04395 + 0.11719

    = 0.171887

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