Answer: (iv) $8/3 \approx 2.67$

Explanation:

Merge of two sorted lists $A=[a_1<a_2]$, $B=[b_1<b_2]$ with all 4! / (2!2!) = 6 interleavings equally likely (distinct elements, random order). Standard merge:

 i=j=0, comps=0
 while i<2 and j<2: comps++; if A[i]<B[j] i++ else j++

Best case = 2 comps (one list entirely smaller: $a_1<a_2<b_1<b_2$ or $b_1<b_2<a_1<a_2$)
Worst case = 3 comps (interleaved).

Enumerate:

1. $a_1<a_2<b_1<b_2$: $a_1\!<\!b_1$, $a_2\!<\!b_1$ → 2
2. $a_1<b_1<a_2<b_2$: $a_1\!<\!b_1$, $b_1\!<\!a_2$, $a_2\!<\!b_2$ → 3
3. $a_1<b_1<b_2<a_2$: $a_1\!<\!b_1$, $b_1\!<\!a_2$, $b_2\!<\!a_2$ → 3
4. $b_1<b_2<a_1<a_2$: 2 (symmetric to 1)
5. $b_1<a_1<b_2<a_2$: 3
6. $b_1<a_1<a_2<b_2$: 3

Average = $(2+3+3+2+3+3)/6 = 16/6 = 8/3$.

Worst-case merging $m+n-1 =3$, best $\min(m,n)=2$, average $8/3$.

Correct choice: (iv) $8/3$