Question

Find the probability of of getting (I)9 (ii)7 and (iii)6 on throwing two dice.

Answer 1

(i) Total number of cases = 6*6 = 36

Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4

So probability = 4/36 = 1/9

(ii) Total number of cases = 6*6 = 36

• There are 6 combinations: (1,6), (6,1), (2,5), (5,2), (3,4) and (4,3).

⇒P(sum=7)=6/36 = 1/6

(iii) Total number of cases = 6*6 = 36

Of these the outcomes (1,5), (2,4), (3,3), (4,2) and (5,1) denote we have a got a sum of 6

There are 5 combinations = Therefore P(sum=6) = 5/36