Question

Two dice are thrown at the same time. Find the probability of getting different number on both dice.

Answer 1

Solve for the required probability:

We know that probability of an event $P\left(E\right)=\frac{n\left(E\right)}{n\left(s\right)}$ where $n\left(E\right)$ and $n\left(s\right)$ are the number of favorable outcomes and total possible number of outcomes respectively.

Given that two dice are thrown $\Rightarrow n\left(s\right)=6^2=36$ [since a dice has $6$ faces]

Number of outcomes for getting same number on both dice i.e. $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)$ is $n_1=6$

Now the favorable number of outcomes i.e. different number on both dice is $n\left(E\right)=n\left(s\right)-n_1$

$\begin{array}{cccl}\Rightarrow & n\left(E\right)& =& 36-6\\ \Rightarrow & n\left(E\right)& =& 30\end{array}$

$\therefore$Required probability $P\left(E\right)=\frac{n\left(E\right)}{n\left(s\right)}$

$\begin{array}{cc}=& \frac{30}{36}\\ =& \frac{5}{6}\end{array}$

Hence the required probability is $\frac{5}{6}$