So far, I have concluded that my sample space obviously ranges from 0-9. I know that the digits can be chosen at random with replacement. Now the outcome probability of each digit being chosen is 1/9. However, I am having trouble figuring out the outcome probability of the entire set. Usually, you compute the event probability for the original question by summing the probability of each individual event. (ie. $1/9 + 1/9 + 1/9...)$ But I am not sure if that applies here, even though it is uniform. Could someone tell me if I am in the wrong direction?
2026-02-23 02:15:30.1771812930
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Out of r digits chosen at random, what is the probability that no digits exceed k. (Range: $0 \le k \le9$.)
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$$P[\text{no digit from the chosen $r$ digits exceeds $k$}]=p^r$$
where $p$ is the probability that a digit doesn't exceed $k$, which is given by
$$p=\sum_{i=0}^kP[\text{digit}=i]=\frac{k+1}{10}$$
where $P[\text{digit}=i]=\frac{1}{10}$ (and not $\frac{1}{9}$ because you have $10$ digits from $0$ to $9$). So, the final answer is $$\left(\frac{k+1}{10}\right)^r$$
"Usually, you compute teh event probability for the original question by summing the probability of each individual event"
This is only done in certain very restrictive scenarios. The property you are thinking of is the following:
Breaking apart a probability as a summation is again only valid when talking about "at least one of these things happening" and specifically when those things can not happen simultaneously. The more generalized version of the above is the following:
In your problem however, we are not looking for "at least one thing happening" but rather we are looking for "all things happening." We want every digit chosen to be less than or equal to $k$.
When we have two events $A$ and $B$ and we want them both to happen simultaneously, this is referring to $Pr(A\cap B)$. Similar to the addition rule above where we can in some circumstances change a $\cup$ inside a probability into a $+$ outside of it, there is a way to change $\cap$ into $\times$, however again like the addition rule above there are only certain circumstances where you are allowed to do this.
This works if and only if the events are independent however, and the usual way to know if events are independent is by manually checking that $Pr(A\cap B)=Pr(A)\times Pr(B)$ or other equivalent conditions. Thankfully most problems will either be worded in a way or use examples that are obviously independent or the problem will directly tell you they are independent. If it is not obvious, then you may have to make assumptions about the independence of the events. The generalized result is the following:
The above properties can of course be extended to arbitrarily many sets.
In your problem, we (presumably) select $r$ digits chosen uniformly and independently at random (the phrase "uniformly and independently" should have been included in the problem statement for there to be an actual way of answering). Now... let $A_1$ be the event that the $1$'st digit is less than or equal to $k$. Similarly define $A_2,A_3,\dots,A_r$.
You should be able to see that $Pr(A_1)=\dfrac{k+1}{10}$ since there are ten equally likely options for the $1$'st digit to be and of those each of $0,1,2,\dots,k$ are less than or equal to $k$, giving $k+1$ "good" results. Each of $Pr(A_2),Pr(A_3),\dots$ can be calculated the same way.
Now, the event that every digit is less than or equal to $k$ is $Pr(A_1\cap A_2\cap \cdots \cap A_r)$ which can be expanded as:
$$\begin{array}{ll}Pr(A_1\cap A_2\cap A_3\cap \cdots \cap A_r)& = Pr(A_1)\times Pr(A_2)\times \cdots Pr(A_r)\\&=\dfrac{k+1}{10}\times \dfrac{k+1}{10}\times \cdots \times \dfrac{k+1}{10}\\&=\left(\dfrac{k+1}{10}\right)^r\end{array}$$
again, using the fact that these events are (assumed to be) independent.