Question

Dice has 6 faces. Four such dice are thrown simultaneously. Find the probability that all of them show the same face.

Answer 1

Total outcomes = 6×6×6×6=1296

All of them show the same faces = (1,1,1,1) (2,2,2,2) (3,3,3,3) (4,4,4,4) (5,5,5,5) (6,6,6,6)=6

P = 6/1296 =1/216

Answer 2

Answer: The required probability is $\dfrac{1}{216}$

Step-by-step explanation:

Dice have 6 faces

Now Four die are thrown simultaneously

Therefore the total outcomes is given by  $6^n$ where n is the number of dice

So the total number of outcomes = $6^4= 1296$

Now Let

E is event such that

E: All of them show the same faces = (1,1,1,1),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)

Now

$P(E)=\dfrac{\text { No. of favourable outcomes }}{\Text { Total outcomes}}\\\\\Rightarrow P(E)= \dfrac{6}{1296 } =\dfrac{1}{216}$

Hence, the required probability is $\dfrac{1}{216}$