Who will check your paper?

There are $7^4$ ways of allocating the teachers to the scripts. If exactly $3$ teachers do the marking, one must mark two scripts, and the other two one script each. There are $7\times {4 \choose 2} = 42$ ways of choosing which two of the four scripts will be marked by the same teacher, and determining which teacher does this marking. There are then $6$ ways of choosing the marker for the third script, and $5$ ways of choosing the marker for the final script. Thus the probability is $\frac{42 \times 6 \times 5}{7^4} \; = \; \frac{5 \times 6^2}{7^3}\;$ making $k = \boxed{\tfrac57}$ .

Total no. of ways in which 4 papers can be checked = $7*7*7*7$ $(\because$ Each paper can be checked by any of the 7 teachers.)

For no. of favourable cases(when all 4 papers are checked by exactly 3 teachers), first of all we select the 3 teachers out of 7 in $\binom{7}{3}$ ways.

Now, we consider the possible no of ways in which 4 papers can be checked by these 3 teachers. Possible no of ways= $3*3*3*3$

This will include the following cases(so they must be subtracted)-

i) All 4 papers are checked by exactly 2 teachers out of 3 chosen

ii)All 4 papers are checked by 1 teacher (3 such cases exist)

No of ways in which papers are checked by 2 teachers= $\binom{3}{2}*2*2*2*2$ (first we selected 2 teachers out of 3) This also includes the case when papers are checked by only 1 teacher but this case is included 2 times here. Hence, no. of ways in which papers are checked by exactly 2 teachers out of 3 chosen= $\binom{3}{2}*2*2*2*2 - 6$

$\therefore$ No of favourable cases = $\binom{7}{3}*\left (3^{4}-\left (\binom{3}{2}*2^4 - 6 \right )-3\right )$ = $35*36$

Hence, required probability= $\frac{35*36}{7*7*7*7}=\frac{5}{7}*\left ( \frac{6}{7} \right )^{2}$ So, k= $\frac{5}{7}=\boxed{0.714285}$