1. Interpretation of the random variable

We toss an unbiased coin $5$ times.

The variable $X$ is defined as:

• $X=3$ if the outcome contains exactly/assigned for $3$ consecutive heads,
• $X=4$ if it contains $4$ consecutive heads,
• $X=5$ if it contains $5$ consecutive heads,
• otherwise $X=-1$.

Since there are only $5$ tosses, the relevant runs of consecutive heads are of lengths $3,4,5$.

A natural interpretation is that $X$ equals the maximum length of a consecutive run of heads when that maximum is $3,4,$ or $5$; otherwise $X=-1$.

Total number of outcomes: $2^5=32.$

2. Find outcomes with maximum run length $5$

This happens only for: $HHHHH$ So, $P(X=5)=\frac{1}{32}.$

3. Find outcomes with maximum run length $4$

This means there is a run of $4$ heads, but not $5$ heads.

Possible sequences:

• $HHHHT$
• $THHHH$

So, $P(X=4)=\frac{2}{32}=\frac{1}{16}.$

4. Find outcomes with maximum run length $3$

This means there is a run of exactly $3$ consecutive heads, but no run of $4$ or $5$ heads.

List such sequences:

• $HHHTT$
• $THHHT$
• $TTHHH$
• $HHHTH$
• $HTHHH$

Thus, $P(X=3)=\frac{5}{32}.$

5. Find probability that no run of $3$ or more heads occurs

Then $X=-1$.

So remaining outcomes: $32-(1+2+5)=24.$ Hence, $P(X=-1)=\frac{24}{32}=\frac{3}{4}.$

6. Compute expectation

$E[X]=3\cdot P(X=3)+4\cdot P(X=4)+5\cdot P(X=5)+(-1)\cdot P(X=-1).$

Substitute values: $E[X]=3\cdot \frac{5}{32}+4\cdot \frac{2}{32}+5\cdot \frac{1}{32}-1\cdot \frac{24}{32}.$

$E[X]=\frac{15+8+5-24}{32}=\frac{4}{32}=\frac{1}{8}.$

7. Conclusion

$\boxed{E[X]=\frac{1}{8}}$

So the correct option is C.