Step 1

Given data:

The probability of 0 bit is \(\displaystyle{P}{\left({0}\right)}={0.6}\).

The expression for the probability of 1 bit is,

\(\displaystyle{P}{\left({1}\right)}={1}-{P}{\left({0}\right)}\)

Substitute the given values in the above expression.

\(\displaystyle{P}{\left({1}\right)}={1}-{0.6}\)

\(\displaystyle={0.4}\)

a) The expression for the probability of code-word 0101 is,

\(\displaystyle{P}{\left({a}\right)}={P}{\left({0}\right)}\times{P}{\left({1}\right)}\times{P}{\left({0}\right)}\times{P}{\left({1}\right)}\)

Substitute the given and above-calculated values in the expression.

\(\displaystyle{P}{\left({a}\right)}={\left({0.6}\right)}{\left({0.4}\right)}{\left({0.6}\right)}{\left({0.4}\right)}\)

\(\displaystyle={0.0576}\)

Thus, the probability of code-word 0101 is 0.0576.

Step 2

b) The expression for the probability that the code-word contains exactly 2 ones is,

\(\displaystyle{P}{\left({b}\right)}=^{{{4}}}{C}_{{{2}}}{\left({P}{\left({1}\right)}\right)}^{{{2}}}{\left({P}{\left({0}\right)}\right)}^{{{2}}}\)

Substitute the given values in the above expression.

\(\displaystyle{P}{\left({b}\right)}=^{{{4}}}{C}_{{{2}}}{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)

\(\displaystyle={\frac{{{4}!}}{{{\left({2}!\right)}{\left({2}!\right)}}}}{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)

\(\displaystyle={6}\times{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)

\(\displaystyle={0.3456}\)

Thus, the probability that the code-word contains exactly 2 ones is 0.3456.

Given data:

The probability of 0 bit is \(\displaystyle{P}{\left({0}\right)}={0.6}\).

The expression for the probability of 1 bit is,

\(\displaystyle{P}{\left({1}\right)}={1}-{P}{\left({0}\right)}\)

Substitute the given values in the above expression.

\(\displaystyle{P}{\left({1}\right)}={1}-{0.6}\)

\(\displaystyle={0.4}\)

a) The expression for the probability of code-word 0101 is,

\(\displaystyle{P}{\left({a}\right)}={P}{\left({0}\right)}\times{P}{\left({1}\right)}\times{P}{\left({0}\right)}\times{P}{\left({1}\right)}\)

Substitute the given and above-calculated values in the expression.

\(\displaystyle{P}{\left({a}\right)}={\left({0.6}\right)}{\left({0.4}\right)}{\left({0.6}\right)}{\left({0.4}\right)}\)

\(\displaystyle={0.0576}\)

Thus, the probability of code-word 0101 is 0.0576.

Step 2

b) The expression for the probability that the code-word contains exactly 2 ones is,

\(\displaystyle{P}{\left({b}\right)}=^{{{4}}}{C}_{{{2}}}{\left({P}{\left({1}\right)}\right)}^{{{2}}}{\left({P}{\left({0}\right)}\right)}^{{{2}}}\)

Substitute the given values in the above expression.

\(\displaystyle{P}{\left({b}\right)}=^{{{4}}}{C}_{{{2}}}{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)

\(\displaystyle={\frac{{{4}!}}{{{\left({2}!\right)}{\left({2}!\right)}}}}{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)

\(\displaystyle={6}\times{\left({0.4}\right)}^{{{2}}}{\left({0.6}\right)}^{{{2}}}\)

\(\displaystyle={0.3456}\)

Thus, the probability that the code-word contains exactly 2 ones is 0.3456.