Question

$A$ and $B$ throw a pair of dice respectively. If $A$ throws die such that the total of the numbers on the dice is $9$, find $B$'s chance of throwing a higher number than A.

Answer 1

In a throw of pair of dice, total no of possible outcomes$=36(6\times 6)$ which are

$(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)$

$(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$

$(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$

$(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$

$(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$

$(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$

Given, $A$ throws $9$

Let $E$ be the event of getting a number more than $9$ when $B$ throws a pair of dice

$No.$ $of$ $favorable$ $outcomes$ $=6$ $(i.e.,(4,6),(5,5),(5,6),(6,4),(6,5),(6,6\left)\right)$

$P\left(E\right)=\frac{6}{36}=\frac{1}{6}$