In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

$nA=6, nB=5, nA\cap B=1$

$\therefore$Required probability =  $PA\cup B$

 =  $PA+PB-PA\cap B$

=   $\frac{6}{36}+\frac{5}{36}-\frac{1}{36}$ =  $\frac{5}{18}$