probability and rolling a die(uniform) If I rolled 6 different fair six-sided dice,what is the probability that the maximum of the rolls is equal to 5?
My solution so far
Let x =outcome of die~uniform $[1,6]$
$p(X\leq x)=\frac{1}{5}$
 and $F(X)=\frac{x-1}{5}$
Let $y_n$ =maximum of the rolled die
Then $F(y_n)=(\frac{y_n-1}{5})^n$
so $p(y_n=5)=p(y_n \leq 5)-P(y_n\leq 4)=\frac{4}{5}^6*\frac{3}{5}^6=0.215$
correct answer 0.25
 A: Note that the dice is six-sided,
$$P(X=x)=\frac16$$
$$P(X \leq x) = \frac{x}6$$
$$P(Y \leq y) = \left( \frac{y}{6}\right)^6$$
$$P(Y=5)=\left(\frac56 \right)^6-\left(\frac46 \right)^6 \approx 0.2471\approx 0.25$$
A: Probability of getting a max of five on all dice is equal to one minus the probability of getting at least one six, which would be:
$$ 1 - \sum_{k=1}^{6} \binom{6}{k} * \left(\frac{1}{6} \right)^k * \left(\frac{5}{6} \right)^\left(6-k\right) \approx 0.33$$ 
Probability that at least one 5 appears without a 6 should be:
$$\sum_{k=1}^{6} \binom{6}{k} * \left(\frac{1}{6} \right)^k * \left(\frac{4}{6} \right)^\left(6-k\right) \approx 0.2471$$
A: In more technical terms, the possible outcomes of a toss of a die can be modeled by the set $\Omega = \{1, 2, 3, 4, 5, 6\}$.
The fairness of this die can be modeled by the symmetric probability space, $S$, on $\Omega$, namely $S = (\Omega, \mathcal{A} = 2^\Omega, P)$, where, for every $E \subseteq \Omega$, $P(E) = \#E/\#\Omega$.
A sequence of six independent tosses of this die is then modeled by the product probability space $\otimes_{i = 1}^6 S = (\Omega', \mathcal{A}', P')$, in which $\Omega' = \Omega^6$, $\mathcal{A}' = \otimes_{i = 1}^6 \mathcal{A} = 2^{\Omega^6}$, and $P' = \otimes_{i = 1}^6 P$, i.e. $P'$ is the unique probability measure on $(\Omega',\mathcal{A}') = \left(\Omega^6, 2^{\Omega^6}\right)$, which assigns to every event $E \subseteq \Omega^6$ of the form $E = E_1\times E_2 \times \cdots \times E_6$, where $E_1, E_2, \dots, E_6 \subseteq \Omega$, the probability $P'(E) = P(E_1)P(E_2)\cdots P(E_6)$.
The event $E$, whose probability we are interested in calculating, is
$$
E := \left\{(\omega_1, \dots, \omega_6) \in \Omega^6\ |\!:\ \max\{\omega_1, \dots, \omega_6\} = 5\right\}.
$$
Since, for every $d\in \{1, 2, \dots, 6\}$ we have:
$$
d = 5 \iff (d\leq 5) \wedge \neg(d\leq 4),
$$
we may represent $E$ as the difference $E = E_1\setminus E_2$ of the two events
$$
\begin{align}
E_1 &:= \left\{(\omega_1, \dots, \omega_6) \in \Omega^6\ |\!:\ \max\{\omega_1, \dots, \omega_6\} \leq 5\right\}, \\
E_2 &:= \left\{(\omega_1, \dots, \omega_6) \in \Omega^6\ |\!:\ \max\{\omega_1, \dots, \omega_6\} \leq 4\right\}.
\end{align}
$$
Note that $E_2 \subseteq E_1$, and, therefore, $P'(E) = P'(E_2) - P'(E_1)$.
Since, for every $(\omega_1, \dots, \omega_6) \in \Omega^6$ and for every $d\in \{1, 2, \dots, 6\}$,
$$
\max\{\omega_1, \dots, \omega_6\} \leq d \iff (\omega_1 \leq d)\wedge\cdots\wedge(\omega_6 \leq d),
$$
we have
$$
E_1 = \left\{(\omega_1, \dots, \omega_6)\in\Omega^6\ |\!:\ (\omega_1 \leq 5)\wedge\cdots\wedge(\omega_6 \leq 5)\right\} = \{1, 2, \dots, 5\}^6,
$$
and, similarly,
$$
E_2 = \{1, 2, 3, 4\}^6.
$$
We finally calculate the probability of the event $E$.
$$
\begin{align}
P'(E) &= P'(E_1) - P'(E_2) \\
&= P'\left(\{1, \dots, 5\}^6\right) - P'\left(\{1, \dots, 4\}^6\right) \\
&= \left(P\left(\{1, \dots, 5\}\right)\right)^6 - \left(P\left(\{1, \dots, 4\}\right)\right)^6 \\
&= \left(\frac{\#\{1, \dots, 5\}}{\#\Omega}\right)^6 - \left(\frac{\#\{1, \dots, 4\}}{\#\Omega}\right)^6 \\
&= \left(\frac{5}{6}\right)^6 - \left(\frac{4}{6}\right)^6 \\
&= \frac{427}{1728}.
\end{align}
$$
In conclusion, $P'(E) \approx 0.247$, rounded to the nearest thousandth.
