GATE EC 2016 Set 1 — Question 59

Let $p$ be the crossover probability, $p = 0.1$. This means the probability of a bit error during transmission is $p$. The probability of correct transmission is $1-p = 0.9$.The source symbols are transmitted with equal probability: $P(0) = P(1) = 0.5$.Case 1: 0 is transmitted.
Transmitted sequence: 000.
The decoder decides 1 if the number of ones in the received sequence is greater than the number of zeros. This means 2 or 3 ones are received.Probability of receiving 2 ones (and 1 zero): $P(2 \text{ ones}) = \binom{3}{2} p^2 (1-p) = 3 \times (0.1)^2 \times (0.9) = 3 \times 0.01 \times 0.9 = 0.027$.
Probability of receiving 3 ones (and 0 zeros): $P(3 \text{ ones}) = \binom{3}{3} p^3 = (0.1)^3 = 0.001$.So, the probability of error when 0 is transmitted, $P(E|0) = P(2 \text{ ones}) + P(3 \text{ ones}) = 0.027 + 0.001 = 0.028$.Case 2: 1 is transmitted.
Transmitted sequence: 111.
The decoder decides 0 if the number of zeros in the received sequence is greater than the number of ones. This means 2 or 3 zeros are received.Probability of receiving 2 zeros (and 1 one): $P(2 \text{ zeros}) = \binom{3}{2} p^2 (1-p) = 3 \times (0.1)^2 \times (0.9) = 3 \times 0.01 \times 0.9 = 0.027$.
Probability of receiving 3 zeros (and 0 ones): $P(3 \text{ zeros}) = \binom{3}{3} p^3 = (0.1)^3 = 0.001$.So, the probability of error when 1 is transmitted, $P(E|1) = P(2 \text{ zeros}) + P(3 \text{ zeros}) = 0.027 + 0.001 = 0.028$.Average probability of error ($P_e$):
$P_e = P(E|0)P(0) + P(E|1)P(1)$
Since $P(0) = P(1) = 0.5$ and $P(E|0) = P(E|1) = 0.028$,
$P_e = 0.028 \times 0.5 + 0.028 \times 0.5 = 0.028$.The average probability of error is $0.028$.