A box contains n balls numbered from 1 to n . The balls are…
A box contains $n$ balls numbered from $1$ to $n$. The balls are drawn at random without replacement. What is the probability that in each of the first $k$ draws the number on the ball coincides with the number of the draw?
Explanation of Method
We need ball $1$ on draw $1$, ball $2$ on draw $2$, …, ball $k$ on draw $k$. By the multiplication rule, the first draw succeeds with probability $\frac1n$, the second (given the first succeeded) with $\frac1{n-1}$, and so on. We check by counting the orders of all balls.
Step 1: Draw by draw
Let $C_i$ = "draw $i$ gives ball $i$".
- $P(C_1)=\frac1n$;
- $P(C_2\mid C_1)=\frac1{n-1}$ (ball $2$ is one of $n-1$ remaining balls);
- …
- $P(C_k\mid C_1\cdots C_{k-1})=\frac1{n-k+1}$.
Step 2: Multiplication rule
\[ \boxed{P=\frac1n\cdot\frac1{n-1}\cdots\frac1{n-k+1}=\frac{(n-k)!}{n!}.} \]
Step 3: Check by counting
All $n!$ orders of the balls are equally likely. The orders starting with $1,2,\dots,k$ number $(n-k)!$ (the remaining balls in any order). So $P=\frac{(n-k)!}{n!}$. ✓
Step 4: Examples
| $n$ | $k$ | $P$ |
|---|---|---|
| $10$ | $1$ | $1/10$ |
| $10$ | $2$ | $1/90$ |
| $10$ | $3$ | $1/720$ |
| $n$ | $n$ | $1/n!$ (all balls in order) |
Note: for $k=n-1$ and $k=n$ the probabilities are equal, $\frac1{n!}$: if the first $n-1$ balls match, the last one must match as well.
Result
| Answer | |
|---|---|
| Probability | $\dfrac1{n(n-1)\cdots(n-k+1)}=\dfrac{(n-k)!}{n!}$ |