Question

54) Two persons each makes a single throw with a pair of dice, The probability that the throws are unequal given by.
$$\begin{gathered} \left(a\right)\frac{1}{6^3} \\ \left(b\right)\frac{73}{6^3} \\ \left(c\right)\frac{51}{6^3} \\ \left(d\right)noneofthese \end{gathered}$$

Answer 1

Hi,
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