# A and B play a dice game using two dice namely 'X' and 'Y'. The numbers inscribed on the six faces of 'X' are 1, 2, 3, 4, 5 and 7, and the numbers inscribed on the six faces of 'Y' are 2, 3, 4, 5, 6 and 8. In each round of the game, each players rolls both the dice simultaneously and records the product of the two numbers appearing on the top of the two dice as his scores for that round. In a particular round, the sum of the scores of A and B is an even number, then how many distinct scores A could have had in that round?

1
by TarynVora785

2015-09-21T01:48:52+05:30

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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

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