A person has 8 keys, only one of which opens a certain door. The person tries the keys at random, without replacement. We need to find the probability distribution of the number of keys required to open the door. Let $X$ be the number of keys needed to open the door.

1. Problem Description

A person has 8 keys, only one of which opens a certain door. The person tries the keys at random, without replacement. We need to find the probability distribution of the number of keys required to open the door. Let $X$ be the number of keys needed to open the door.

2. Solution Steps

The random variable $X$ can take values from 1 to

8. The probability that the door opens on the first try is $P(X=1) = \frac{1}{8}$.

The probability that the door opens on the second try is $P(X=2) = \frac{7}{8} \times \frac{1}{7} = \frac{1}{8}$. This is because the first key must be incorrect (probability $\frac{7}{8}$), and then the second key must be the correct one out of the remaining 7 keys (probability $\frac{1}{7}$).

The probability that the door opens on the third try is $P(X=3) = \frac{7}{8} \times \frac{6}{7} \times \frac{1}{6} = \frac{1}{8}$.

In general, the probability that the door opens on the $k$-th try, where $1 \le k \le 8$, is

$P(X=k) = \frac{7}{8} \times \frac{6}{7} \times \frac{5}{6} \times \dots \times \frac{8-(k-1)}{8-(k-2)} \times \frac{1}{8-(k-1)} = \frac{8-k+1}{8} \times \frac{1}{8-k+1} = \frac{1}{8}$.

Therefore, $P(X=k) = \frac{1}{8}$ for $k=1, 2, ..., 8$.

The probability distribution is:

$P(X=1) = \frac{1}{8}$

$P(X=2) = \frac{1}{8}$

$P(X=3) = \frac{1}{8}$

$P(X=4) = \frac{1}{8}$

$P(X=5) = \frac{1}{8}$

$P(X=6) = \frac{1}{8}$

$P(X=7) = \frac{1}{8}$

$P(X=8) = \frac{1}{8}$

3. Final Answer

The probability distribution of the number of keys needed to open the door is:

$P(X=k) = \frac{1}{8}$ for $k=1, 2, 3, 4, 5, 6, 7, 8$.