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54) Two persons each makes a single throw with a pair of dice, The probability that the throws are unequal given by.

$\left(a\right)\frac{1}{{6}^{3}}\phantom{\rule{0ex}{0ex}}\left(b\right)\frac{73}{{6}^{3}}\phantom{\rule{0ex}{0ex}}\left(c\right)\frac{51}{{6}^{3}}\phantom{\rule{0ex}{0ex}}\left(d\right)noneofthese$

P(scores unequal) = 1 - P(equal) = 1 - P(both total 2 or both total 3 or . . . or both total 12)

P(both total 2) = P(1st totals 2)*P(2nd total 2) = (1/36)*(1/36) = 1/1296

P(both total 3) = P(1st totals 3)*P(2nd totals 3) = (2/36)*(2/36) = 4/1296

In a similar way the probability that both total 4, 5, 6, 7, 8, 9, 10, 11, 12 are

9, 16, 25, 35, 25, 16, 9, 4, 1 each over 1296

P(scores equal) = (1 + 4 + 9 + 16 + 25 + 36 + 25 + 16 + 9 + 4 + 1)/1296

= 146/1296 = 73/648

P(scores unequal) = 1 - 73/648 = 575/648

Regards

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