Two different dice are thrown together find the probability that the product of the numbers is 18

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