The average bit error rate at the input of a (7, 4, 1) Hamming decoder is 0.10. The probability that the decoder will fail to decode a received word correctly is _______. (rounded off to two decimal places)

The correct answer is 0.15.

Let us analyze the problem step-by-step to understand why the probability of a decoding failure is 0.15.

A (7, 4, 1) Hamming code has the following parameters:
- Codeword length ($n$) = 7 bits
- Number of message bits ($k$) = 4 bits
- Error correction capability ($t$) = 1 bit

This means that the Hamming decoder can detect and correct any single-bit error in a received 7-bit codeword. If there are 0 or 1 errors in the received word, the decoder will decode it correctly. However, if there are 2 or more errors in the 7-bit codeword, the decoder will fail to decode the received word correctly.

The average bit error rate (probability of a single bit being received in error) is given as:
$p=0.10$

Assuming bit errors occur independently, the number of errors in a 7-bit codeword follows a binomial distribution. The probability of having exactly $i$ errors in a 7-bit word is given by:

$P(ierrors)=\left(\genfrac{}{}{0ex}{}{7}{i}\right)p^i{(1-p)}^{7-i}$

The decoder will decode the word correctly if there are 0 or 1 errors. Let $P_c$ be the probability of correct decoding:

$P_c=P(0errors)+P(1error)$

Let us calculate these two probabilities:
For 0 errors:

$P(0errors)=\left(\genfrac{}{}{0ex}{}{7}{0}\right){\left(0.10\right)}^0{\left(0.90\right)}^7=1\times 1\times {0.90}^7$

$P(0errors)\approx 0.4783$

For 1 error:

$P(1error)=\left(\genfrac{}{}{0ex}{}{7}{1}\right){\left(0.10\right)}^1{\left(0.90\right)}^6=7\times 0.10\times {0.90}^6$

$P(1error)\approx 7\times 0.10\times 0.531441=0.3720$

Now, calculate the total probability of correct decoding:

$P_c\approx 0.4783+0.3720=0.8503$

The probability that the decoder fails to decode the received word correctly ($P_f$) is the complement of the probability of correct decoding:

$P_f=1-P_c$

$P_f\approx 1-0.8503=0.1497$

Rounding off to two decimal places, we get:

$P_f\approx 0.15$