Find the chance of throwing at least one ace in a single throw with two dice.

Hint: We will take total sample space in two dice is $36$. Then, we will calculate the one ace. By using formula of $P(E) = \dfrac{{favourable\,\,outcome}}{{total\,\,number\,\,of\,\,outcome}}$


Complete step by step solution:
As we know that sample space in a single dice $\{ 1,2,3,4,5,6) = 6$
So, total number of outcomes in two dice $ = {6^2}$
total number of outcomes in two dice $ = 36$
Now, we make sample space
\[\begin{array}{*{20}{c}}
  {(1,1)}&{(1,2)}&{(1,3)}&{(1,4)}&{(1,5)}&{(1,6)} \\
  {(2,1)}&{(2,2)}&{(2,3)}&{(2,4)}&{(2,5)}&{(2,6)} \\
  {(3,1)}&{(3,2)}&{(4,3)}&{(3,4)}&{(3,5)}&{(3,6)} \\
  {(4,1)}&{(4,2)}&{(4,3)}&{(4,4)}&{(4,5)}&{(4,6)} \\
  {(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}&{(5,5)}&{(5,6)} \\
  {(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&{(6,5)}&{(6,6)}
\end{array}\]
Here $\{ (1,2),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)\} $at least one ace.
And rest have no ace in dice $ = 25$
So, at least one ace $ = 11$=
Now, by using the formula of probability,
$P($at least one dice in a single throw)$ = \dfrac{{favourable\,\,outcome}}{{total\,\,number\,\,of\,\,outcome}}$
Here, favourable outcomes $ = 11$
and total number of outcome $ = 36$
Then
P(at least one ace in a single throw)$ = \dfrac{{11}}{{36}}$


Additional Information: We can answer this question in another way. Here out of the sample space, there will be $25$outcomes where No. ace
$
  \{ (2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,5),(4,6),(5,2),(5,3),(5,4),(5,5),(5,6), \\
  (6,2),(6,3),(6,4),(6,5),(6,6)\} \\
 $
So, $P(no.\,ace) = \dfrac{{favourable\,\,outcome}}{{total\,\,number\,\,of\,\,outcome}}$
$P$(no. ace)$ = \dfrac{{25}}{{36}}$
The probability that there will be at least one ace $ = 1 - \dfrac{{25}}{{36}}$
One ace $ = \dfrac{{36 - 25}}{{36}}$
One ace $ = \dfrac{{11}}{{36}}$
Hence, the probability of at least one ace in a single throw $ = \dfrac{{11}}{{36}}$


Note: Students must know that if the first dice is an ace, it does not matter what the other dice is. So, there are $6$combinations of the $36$. If the second dice is an ace, it does not matter what the first dice is. Then $6 + 6 - = 11$