I guess that in the above exercise, in the last sentence, there is some mistake. That should be the number of distinct scores that A and B could have had in a round. That is score A ≠ score B.

let SA mean Score(A) in a round. SB =score(B). We know the conditional probability by Beyes' theorem.

P( event M | event N) = P( M Π N ) / P(N)

X: 1,2,3,4,5,7 Y: 2,3,4,5,6,8

SA or SB = odd when score = 3,5,9,3*5,5*3,25,21,35

Prob[SA = odd] = P[SB = odd] = 8/36

P(SA = even] = P[ B = even] = 28/36

P(N) = P(Sum SA + SB = even)

= P(SA = even) * P(SA = even) + P(SA=odd) * P(SB=odd)

= (28² + 8² ) / 36² = *53 / 81*

We will now find the P( M Π N) = P(sum = even Π SA ≠ SB)

1) when SA = odd and SB = odd,

then score 15 occurs in two combinations. So when SA = 15, SB can only be the other 6 combinations. When SA ≠15, then SB can be any other other 7 scores.

** P(SA= odd Π SB = odd Π A ≠ B) = (6 * 7 + 2 * 6) / 36² = 1/24**

2) When SA = even and SB = even.

(a) the scores 6, 8, 12 are obtainable in 3 combinations each. (1*6,2*3,3*2; 1*8, 2*4, 4*2 ; 2*6, 3*4, 4*3). If SA = 6, 8 or 12, then SB can be one of 25 other even scores. Prob. = 9 * 25 / 36² = 25/144

(b) the scores 4, 10, 16, 20 ,24 are obtainable in 2 combinations each.

(1*4, 2*2, 2*5, 5*2, 2*8, 4*4, 4*5, 5*4, 4*6, 3*8)

if SA = any of above, then SB can be any of other even 26 scores.

Prob = 10 * 26 / 36² = 65 / 324

(c) the remaining (28 - 9 - 10) = 9 scores are obtainable only 1 way. When SA = any of them, SB can be any of the remaining 27 even scores.

Prob = 9 * 27 / 36² = 3/16

** P(SA = even Π SA = even Π SA ≠ SB) = 25/144 + 65/324 + 3/16 = 91/162**

P(Sum = even ** Π** SA ** ≠ SB)** = 1/24 + 91/162 = 391 / 648

==

*Finally, the required Prob. P[ Score A ≠ Score B | Sum scores = Even) *

* = [391 / 648] / [53/81] = 391 / 424 = 0.922*