Question

Two different dice are thrown together. Find the probability that
(i) the sum of the numbers appeared is less than 7.
(ii) the product of the numbers appeared is less than 18.

Answer 1

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 = $\left[\right(1,1);(1,2);(1,3);(1,4);(1,5);(1,6);(2,1);(2.2);(2,3);(2,4);(2,5);(2,6);(3,1);(3,2);(3,3);(3,4);(3,5);(3,6);(4,1);(4,2);(4,3);(4,4);(4,5);(4,6);(5,1);(5,2);(5,3);(5,4);(5,5);(5,6);(6,1);(6,2);(6,3);(6,4);(6,5);(6,6\left)\right]$

n(s) = 36

(i) A : the sum of the numbers appeared is less than 7.

Favourable outcomes: $(1,1);(1,2);(1,3);(1,4);(1,5);(2,1);\left(2.2\right);(2,3);(2,4);(3,1);(3,2);(3,3);(4,1);(4,2);(5,1)$

n(A) = 15

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$

(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

$p\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{26}{36}=\frac{13}{18}$