You’re looking for what’s called the binomial distribution, which gives the probability of exactly $k$ successes in a sequence of $n$ independent Bernoulli (i.e., succeed/fail) trians with a fixed probability $p$ of success. The formula is
$${n \choose k}p^k(1-p)^{n-k}.$$
For your specific problem involving dice, $p$ would be the probability of rolling a one on a single die, i.e., $\frac16$ for a d6 and $\frac1{10}$ for a d10, $n$ would be the total number of dice you’re rolling, and $k$ is the number of ones rolled.
To combine different values of $k$, you add up their respective probabilities. There’s a special name for the sum of the probabilities from $k=0$ to $m$: the cumulative probability distribution, often written $P(X\le m)$. There are formulas for it that you can look up on the web, but I’m not convinced that for small numbers of dice they’re any easier to use than computing the individual probabilities and adding them up yourself. You’re likely not interested in including the probability of not rolling any ones at all in your sums, so you’d need to subtract $P(0)$ from the cumulative probability to get $P(1\le X\le m)$.
When doing these sums, you can save yourself some work by taking advantage of the recurrence $$p(n-k)P(k)=(1-p)(k+1)P(k+1)$$ or, rearranged, $$P(k+1)=\frac{p}{1-p}\frac{n-k}{k+1}P(k)$$ Note that $P(0)=(1-p)^n$.
Let’s work through your example of rolling ones on 4d10 ($n=4, p=1/10, p/(1-p)=1/9$): $$\begin{align}
P(0) &= (1-\frac1{10})^4 = 0.6561 \\
P(1) &= \frac19\cdot\frac41\cdot P(0)=0.2916 \\
P(2) &= \frac19\cdot\frac32\cdot P(1)=0.0486 & P(1\le X\le2)=0.3402 \\
P(3) &= \frac19\cdot\frac23\cdot P(2)=0.0036 & P(1\le X\le3)=0.3438 \\
P(4) &= \frac19\cdot\frac14\cdot P(3)=0.0001 & P(1\le X\le4)=0.3439
\end{align}$$
As a sanity check, the last cumulative value is $1-P(0)$ as expected.
If you want to compute the probabilities for rolling less than or equal to some value, then you adjust the value of $p$. E.g., to find the probability of rolling 3 or less on $k$ dice when rolling 4d10, you’d set $n=4$ and $p=3/10$ in the formula.
you have done it correctly.
Even if two dice are indistinguishable the probability of getting 3 from one die and 4 from another will remain same.
we can not distinguish between (3,4) and (4,3) but it does not affect the fact that they can occur in two different ways.
Best Answer
Outline:
For a single roll of the die the mean number of points is $\mu = 3.5$ and the variance of the number of points is $\sigma^2 = 2.916667.$ These values can be found using the definitions of the mean and variance of a discrete random variable.
Let $T_n$ be the total points on $n$ independent rolls of the die. Then $E(T_n) = n\mu,$ $Var(T_n) = n\sigma^2,$ and $SD(T_n) = \sigma\sqrt{n}.$
According to the CLT, You want $$P(T_n \ge 4500) = P\left(\frac{T_n -n\mu}{\sigma\sqrt{n}}\ge\frac{4500 -n\mu}{\sigma\sqrt{n}}\right)\\ \approx P\left(Z\ge\frac{4500 -n\mu}{\sigma\sqrt{n}} = 1.96 \right)= 0.975.$$
In the next-to-last member of the above equation, use known values of $\mu$ and $\sigma,$ solve for $n$ and round up to the next higher integer.
Note: On account of the following simulation in R, I'm guessing $n$ is not far from $1320.$