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. 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 |