Question

Two dice are thrown, find the probability of getting
(1) The sum of numbers on their upper faces is at least $10$.
(2) The sum of the numbers on their upper faces is divisible by $5$.
(3) The number on the upper face of the first die is greater than the number on the upper face of the second die.

Answer 1

Two dice are thrown
$\therefore$ Sample space

$S=\left\{\right(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6\left)\right\}$
$\therefore n\left(S\right)=36$

(1) $A=\left\{\right(4,6),(5,5),(5,6),(6,4),(6,5),(6,6\left)\right\}$
$n\left(A\right)=6$
Probability, $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$

(2) $B=\left\{\right(1,4),(2,3),(3,2),(4,1),(4,6),(5,5),(6,4\left)\right\}$
$n\left(B\right)=7$
Probability, $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{7}{36}$

(3) $C=\left\{\right(2,1),(3,1),(3,2),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5\left)\right\}$
$n\left(C\right)=15$
Probability, $P\left(C\right)=\frac{n\left(C\right)}{n\left(5\right)}=\frac{15}{36}$