Three identical dice are rolled. The probability that the same number will appear on each of them, is 

Solution:

Let $A_i$ denote an event of getting number  $i(i=1,2,\dots ,6)$ on each die. Then, $A_i,i=1,2,\dots ,6$ are mutually exclusive events.

$\therefore$Required probability $=P\left(A_1\right)+P\left(A_3\right)+D(A,)$    …(i)

Now,

$P\left(A_i\right)=$ Probability of getting number ion each die 

$\Rightarrow P\left(A_i\right)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{216}$

So, required probability $=\frac{6}{216}=\frac{1}{36}$                  $\text{ [From (i)] }$

$\underline{ALITER}$ We have

Total number of elementary events$= 6 x 6 x 6 = 216$

Same number can be obtained on each dice in one of the following ways: 

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

Favourable number of elementary events $= 6$
Hence, required probability $=\frac{6}{216}=\frac{1}{36}$