Question

A pair of fair dice is tossed. Find the probability of getting​  at most a total of 10.

Answer 1

Solution

Total number of cases when two dice roles $=6×6=36$ and are as shown.

$\begin{matrix} & 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 1+1 & 1+2 & 1+3 & 1+4 & 1+5 & 1+6 \\ 2 & 2+1 & 2+2 & 2+3 & 2+4 & 2+5 & 2+6\\ 3 & 3+1 & 3+2 & 3+3 & 3+4 & 3+5 & 3+6 \\ 4 & 4+1 & 4+2 & 4+3 & 4+4 & 4+5 & 4 +6 \\ 5 & 5+1 & 5+2 & 5+3 & 5+4 & 5+5 & 5+6 \\ 6 & 6+1 & 6+2 & 6+3 & 6+4 & 6+5 & 6+6 \end{matrix}$

Let A be the event that on throw of the pair of fair dice, getting a total of more than 10

$A=[(5+6),(6+5),(6,6)]$

$P(A) =\dfrac{3}{36}$

$P(at most a 10) = 1-A$

$P(at most a 10) =\dfrac{33}{36}=\dfrac{11}{12}$