P(X(t) > 2) = Q(2) = 1 - Φ(2) ≈ 0.023
E[Y(t)] = E[X(t)] * |H(0)| = 0
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A source generates a random sequence of bits (0s and 1s) with a probability of 0.6 for a 1 and 0.4 for a 0. What is the probability that a single bit is in error when transmitted over a noisy channel with a probability of error 0.1? P(X(t) > 2) = Q(2) = 1 - Φ(2) ≈ 0
The probability that X(t) > 2 is given by: P(X(t) > 2) = Q(2) = 1 - Φ(2) ≈ 0
was added more content to make your researching a lot easier P(X(t) > 2) = Q(2) = 1 - Φ(2) ≈ 0