When a pair of dice are thrown, then total no. of possible outcomes = 6 ×6 = 36

let E ⟶event of getting sum on dice greater than 10

then no. of favourable outcomes = 3 {(5, 6) (6, 5) (6, 6)}

we know that, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

i.e., P(E) = 3/36 = 1/12