Answer : ` [i] 5/12 [ii] 13/18`
Solution : [i] Number of all possible outcomes is 36.
Let E be the event of getting the sum less than 7 on the two dice.
Then, the favourable outcomes are
[1,1],[1,2],[1,3],[1,4],[1,5],[2,1],[2,2],[2,3],[2,4],[3,1],[3,2],[3,3],[4,1],[4,2],[5,1].
Number of favourable outcomes = 15.
` :. P[E ] = 15/36 = 5/12`.
Solution
If two different dice are thrown together, they have numbers 1, 2, 3, 4, 5 and 6 and 1, 2, 3, 4, 5 and 6 on them.
Total number of outcomes -
S = [[1,1];[1,2];[1,3];[1,4];[1,5];[1,6];[2,1];[2.2];[2,3];[2,4];[2,5];[2,6];[3,1];[3,2];[3,3];[3,4];[3,5];[3,6];[4,1];[4,2];[4,3];[4,4];[4,5];[4,6];[5,1];[5,2];[5,3];[5,4];[5,5];[5,6];[6,1];[6,2];[6,3];[6,4];[6,5];[6,6]]
n[s] = 36
[i] A : the sum of the numbers appeared is less than 7.
Favourable outcomes: [1,1];[1,2];[1,3];[1,4];[1,5];[2,1];[2.2];[2,3];[2,4];[3,1];[3,2];[3,3];[4,1];[4,2];[5,1]
n[A] = 15
P[A]=n[A]n[S]=1536=512
[ii] B: the product of the numbers appeared is less than 18.
Favourable outcomes: [1,1];[1,2];[1,3];[1,4];[1,5];[1,6];[2,1];[2,2];[2,3];[2,4];[2,5];[2,6];[3,1];[3,2];[3,3];[3,4];[3,5];[4,1];[4,2];[4,3];[4,4];[5,1];[5,2];[5,3];[6,1];[6,2]
n[B] = 26
p[B]=n[B]n[S]=2636=1318