A person has 8 keys, only one of which opens a door. They try the keys at random, excluding the used keys from further selection. We need to find the probability distribution of the number of keys needed to open the door. Let $X$ be the random variable representing the number of keys tried until the correct key is found. $X$ can take values from 1 to 8.

1. Problem Description

A person has 8 keys, only one of which opens a door. They try the keys at random, excluding the used keys from further selection. We need to find the probability distribution of the number of keys needed to open the door. Let $X$ be the random variable representing the number of keys tried until the correct key is found. $X$ can take values from 1 to

8.

2. Solution Steps

We want to find $P(X=k)$ for $k = 1, 2, ..., 8$.

- $P(X=1)$: The probability that the first key tried is the correct key. Since there is only one correct key out of 8, $P(X=1) = \frac{1}{8}$.

- $P(X=2)$: The probability that the first key is incorrect and the second key is correct. The probability that the first key is incorrect is $\frac{7}{8}$. Given that the first key is incorrect, there are 7 keys remaining, and only 1 is correct. So, the probability that the second key is correct is $\frac{1}{7}$. Therefore, $P(X=2) = \frac{7}{8} \times \frac{1}{7} = \frac{1}{8}$.

- $P(X=3)$: The probability that the first two keys are incorrect and the third key is correct. The probability that the first key is incorrect is $\frac{7}{8}$. Given that the first key is incorrect, the probability that the second key is incorrect is $\frac{6}{7}$. Given that the first two keys are incorrect, the probability that the third key is correct is $\frac{1}{6}$. Therefore, $P(X=3) = \frac{7}{8} \times \frac{6}{7} \times \frac{1}{6} = \frac{1}{8}$.

- $P(X=k)$: In general, the probability that the first $k-1$ keys are incorrect and the $k$-th key is correct is:

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

$P(X=k) = \frac{7}{8} \times \frac{6}{7} \times ... \times \frac{9-k}{10-k} \times \frac{1}{9-k}$

$P(X=k) = \frac{7 \times 6 \times 5 \times ... \times (9-k)}{8 \times 7 \times 6 \times ... \times (10-k)} \times \frac{1}{9-k}$

$P(X=k) = \frac{9-k}{8} \times \frac{8-k+1}{9-k} \times \frac{7-k+1}{8-k+1} ... \frac{1}{9-k} = \frac{(8-(k-1))!}{(8-(k-2))! \frac{1}{9-k}} = \frac{8!/(9-k)!}{8!/(8-k)!} \frac{1}{(8!/(9-k)!)(8!/(8-k-1)!} = \frac{1}{8}$.

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

3. Final Answer

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