For the first question:
Let's think about just selecting the first digit. There's a probability of $\frac 1 7$ of selecting $7$, $\frac 1 7$ of selecting $6$, and $\frac 1 7$ of selecting $5$. In any of these cases, the resulting number is greater than $4200$. Add up these probabilities: $\frac 1 7 + \frac 1 7 + \frac 1 7 = \frac 3 7$.
Then there's also a $\frac 1 7$ probability of selecting a $4$. Then we need to pick the second digit. If we pick a $2$, then we note that our number will still be greater than $4200$, because picking zeroes for the last two digits is not an option. So the probability of picking one of $\{2,3,5,6,7\}$ is $\frac 5 6$, and any of these will yield a number greater than $4200$. Note that we can't pick $4$ as the second digit because we're selecting from $D$ without replacement.
What's the probability of picking a $4$ and then picking one of $\{2,3,5,6,7\}$? They are independent, so by the multiplication rule we can just multiply them: $\frac 1 7 \cdot \frac 5 6 = \frac 5 {42}$.
So the total probability is $\frac 3 7 + \frac 5 {42} = \frac {18} {42} + \frac {5}{42} = \frac {23}{42}$.
For the second question:
Probability of seeing at least two differently numbered faces $= 1-$
Probability of all four tosses showing the same face.
How do we calculate the probability of all four tosses showing the same face?
Well, suppose we said that all four tosses had to show $6$. The probability of that for a single toss is $\frac 1 6$, and by the rule of multiplication for probabilities of independent events, the chance of tossing four $6$s is $\left(\frac 1 6\right)^4$.
And that's the exact same for all other faces $1,2,3,4,5$. There are six such faces in total, so the probability of all four tosses showing the same face is $6 \cdot \left(\frac 1 6 \right)^4 = 6 \cdot \frac 1 {6^4} = \frac 6 {6^4} = \frac 1 {6^3}$.
(If you want more intuition on this -- draw out the table of probabilities for two rolls of the dice, and note that the diagonal is exactly six events.)
So the probability of seeing at least two numbered faces is $1 - \frac 1 {6^3} = 1 - \frac 1 {216} = \frac {215}{216}$.
The "fast" answer:
The number of 4-digit numbers greater than 4321 using digits in $\{0,1,...,5\}$ is equal to
$$5555_{(6}-4321_{(6}=1234_{(6}=310$$
where the subscript $5555_{(6}$ means that the number is written in base 6. See here for details on numbering in bases (radices) different from 10, and here for details on how to do arithmetic in other bases.
If you are not familiar with non-base-10 numbering:
You can get the same result by counting:
- All numbers that start with $5$: in total $6\cdot 6\cdot 6=216$ numbers.
- All numbers that start with $44$ or $45$: in total $2\cdot 6\cdot 6=72$ numbers.
- All numbers that start with $433$ or $434$ or $435$: in total $3\cdot 6=18$ numbers.
- All numbers tat start with $4322$ or $4323$ or $4324$ or $4325$: in total $4$ numbers.
Grand total:
$$216+72+18+4=310$$
Note that the actual math done in both methods is the same:
$$1234_{(6}=1\cdot6\cdot 6\cdot 6 + 2\cdot 6\cdot 6 + 3\cdot 6 +4 =310$$
Best Answer
Here's a generalization:
The sample space for a six-sided die is $S = \{1,2,3,4,5,6\}$.
Theorem: There are exactly $\color{blue}{3}^n$ n-tuples $(x_{n-1}, x_{n-2}, \cdots, x_1, x_0)$ that satisfy $f(n) = 10^{n-1}x_{n-1} + 10^{n-2}x_{n-2} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^n}$ for $x_i \in S$
Proof by induction:
Base case $n=1$: We have $f(1) = x_0 \equiv 0 \mod{2}$ which is clearly satisfied by exactly $\color{blue}{3}$ 1-tuples $(2), (4)$ and $(6)$.
Inductive step: Assume true for $n$. Now for $n+1$ we have $\begin{array}{l}f(n+1) &= 10^nx_n &+ & 10^{n-1}x_{n-1} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^{n+1}} &\text{(i)}\\&\Rightarrow 0 &+ & 10^{n-1}x_{n-1} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^{n}}\\&\Rightarrow & & 10^{n-1}x_{n-1} + \cdots + 10x_1 + x_0 \equiv 0 \mod{2^{n}}& \text{(ii)}\end{array}$
But $\text{(ii)}$ has exactly $\color{blue}{3}^n$ n-tuple solutions from our inductive hypothesis.
For each such n-tuple $(x_{n-1}, x_{n-2}, \cdots, x_1, x_0)$ , there are $\color{blue}{3}$ unique $x_n \in S$ that satisfy $\text{(i)}$ per Lemma below. So, the number of (n+1)-tuple solutions $(x_n, x_{n-1}, \cdots, x_1, x_0)$ is $\color{blue}{3} \cdot \color{blue}{3}^n = \color{blue}{3}^{n+1}$ and this completes the proof by induction.
Answer:
The probability of forming an n-digit number $x_{n-1}x_{n-2}\cdots x_1x_0 = f(n)$ divisible by $2^n$ is thus $\frac{\text{count of n-digit numbers } \equiv 0 \mod{2^n}}{\text{total count of n-digit numbers}} = \frac{3^n}{6^n} = \frac{1}{2^n}$. In your case, $n=3$, giving the probability of $\frac{1}{8}$.
Lemma: $10^n x + m \equiv 0 \mod{2^{n+1}} $ has exactly $\color{blue}{3}$ unique solutions $x\in S$ given that $m \equiv 0 \mod{2^n}$.
Proof: We have $m \equiv 0 \mod{2^n} \implies m = 2^n\ell$ where $\ell \in \mathbb{Z}$.
$\begin{array}{cl} \therefore &10^n x + m &\equiv 0\mod{2^{n+1}}&\\ \iff & 10^n x + 2^n\ell &= 2^{n+1} q & (q \in \mathbb{Z})\\ \iff & 5^nx + \ell &= 2q &(\text{divide by } 2^n)\\\iff & x + \ell &\equiv 0 \mod{2} & (5^n \equiv 1 \mod{2})\end{array}$
And this has exactly $\color{blue}{3}$ unique solutions $x\in S$ since,
$\mod{2}:\begin{cases}\ell \equiv 0 \implies x + 0 \equiv 0 \implies x \equiv 0 &\implies x \in \{2, 4, 6\} \\ \ell \equiv 1 \implies x + 1 \equiv 0 \implies x \equiv 1 &\implies x \in \{1, 3, 5\}\end{cases}$