We have 6 unbiased dices that are rolled simultaneously.
As, dices are unbiased. Then,
P(x) = $1/6$ where x $\in$ {1, 2, 3, 4, 5, 6}.
When a dice is rolled then any number out of 6 can show up. So, with 1 dice we have 6 possible outcomes.
And, when we have 6 dices and each dices have 6 choices then total possible outcomes = $6^{6}$.
Total cases : $6^{6}$.
Now, we want different numbers on all 6 dices. So, combination like these are our feasible cases :
(1, 2, 3, 4, 5, 6), (1, 3, 5, 2, 4, 6), (2, 4, 6, 5, 3, 1) , (6, 1, 3, 5. 4, 2), - - - - - - - - - - , (6, 5, 4, 3, 2, 1)
How many such feasible case?
It will be 6! = 720
Feasible cases : 6! = 720
So, Required Probability : Feasible cases / Total Cases
= 6! / $6^{6}$.
= 5 / 324.
Hence, Option B is correct Answer.
