This question appears in Stirzaker's Elementary Probability (2nd Edition). It is worked example 1.8 (p40).
My solution and answer:
The answer is $\mathcal{P}(A^c)$, where:
$A = \lbrace (2, 4), (4,2), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4) \rbrace$.
There are 8 pairs, so the probability is $ 1 – 8/36 = 28/36 = 7/9$.
The textbook's solution and answer:
It is routine to list the outcomes that do have a common factor greater than unity. They are 13 in number, namely:
$\lbrace (i,i); i \geq 2 \rbrace, (2, 4), (4,2), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4)$.
This is the complementary event, so by (1.4.5) the required probability is
$ 1 – 13/36 = 23/36 $
Where'd the number 13 come from? Stirzaker enumerates the same set, but says there are 13 members. Where does he get the extra 5?
Screenshot of solution:
Best Answer
The extra five are given by $\{(2,2); (3,3); (4,4); (5,5); (6,6)\}$, which is precisely the set $\{(i,i), i \geq 2\}$.