If $3$ identical Dice are tossed simultaneously, The find probability that all dice shows
same number.
$\bf{My\; Try::}$ Let $A$ be the event in which upper face of all dice shows same number
and $S$ be the sample space
Now Here we have $3$ identical dice.
So $x_{1}$ be the number of times in which dice shows number $1$ on upper face.
Similarly $x_{2}$ be the number of times in which dice shows number $2$ on upper face.
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$x_{6}$ be the number of times in which dice shows number $6$ on upper face.
So here $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6} = 3,$ Where $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}\geq 0$
So we get $\displaystyle n(S) = \binom{8}{2}=56$ and $n(A) = 6$
So required probability $\displaystyle P(A) = \frac{6}{56}$
although i have solve that question using that post
How many are the possible outcomes from throwing $n$ (identical) dice
but i did not understand what is the difference between identical dice and simple dice.
and why answer can not be equal to $\displaystyle \frac{6}{256}$
bcz whether dice are identical or not total number of possible outcome will remain same.)(Its my assumption.)
plz explain me, Thanks
Best Answer
Note that, as the outcomes of the three dice are independent, throwing them simultaneously is equivalent to throwing them after each other. Also note that, for all numbers to be equal, there is no restriction on the outcome of the first die. Assume the outcome of the first die is $n$, with $1\leq{n}\leq6$. Then the probability that all dice show the same number is the probability you throuw $n$ with both the second and the third die, i.e. $$ \frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}. $$