Topic: Communications
What is the average probability of error with two symbols 0 and 1 with source probability 0.2, 0.8 respectively transmitted over a binary asymmetric channel as given below:
(0 -> 0 : 0.7, 0 -> 1 : 0.3)
(1 -> 0 : 0.4, 1 -> 1 : 0.6)
- A. 0.35
- B. 0.38
- C. 0.12
- D. 0.62
Answer: B. 0.38
Explanation: The average probability of error $P_e$ is the sum of the probabilities of transmitting a 0 and receiving a 1, plus transmitting a 1 and receiving a 0.
Given source probabilities: $P(X=0) = 0.2$, $P(X=1) = 0.8$.
Transition probabilities for error: $P(Y=1|X=0) = 0.3$ and $P(Y=0|X=1) = 0.4$.
$P_e = P(X=0)P(Y=1|X=0) + P(X=1)P(Y=0|X=1) = (0.2 \times 0.3) + (0.8 \times 0.4) = 0.06 + 0.32 = 0.38$.