Problem 45 - A First Course in Probability

Question

A woman has \(n\) keys, of which one will open her door. (a) If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her \(k\) th try? (b) What if she does not discard previously tried keys?

Step-by-Step Solution

Answer

(a) If the woman discards the keys after each try, the probability of her opening the door on her \(k\)th try is: \(P(k) = \frac{n - (k -1)}{n}\) (b) If the woman does not discard previously tried keys, the probability of her opening the door on her \(k\)th try is: \(Q(k) = \left(\frac{n-1}{n}\right)^{k-1}\cdot\frac{1}{n}\)

Step 1: Understanding the situation

In this case, the woman discards the keys once they are tested. This implies that she has one fewer key to try after each attempt.

Step 2: Calculate the probability of opening the door on the kth try

First, calculate the probability of failing at each try until the (k-1)th try, and then calculate the probability of success on the kth try. Multiply these probabilities to get the required probability. Let P(k) be the probability of opening the door at the kth try. \(P(k) = \frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdots \frac{n-(k-1)}{n-(k-2)} \cdot \frac{1}{n-(k-1)} \) This simplification can be achieved by canceling out the common factors in the numerator and the denominator.

Step 3: Simplify the result

\(P(k) = \frac{n - (k -1)}{n} \) Therefore, the probability of opening the door on her kth try in this case is: \(\frac{n - (k -1)}{n} \) **(b) When the keys are not discarded after each try**

Step 1: Understanding the situation

In this case, the woman does not discard the keys once they are tested. This implies that she has the same number of keys to try during each attempt.

Step 2: Calculate the probability of opening the door on the kth try

First, calculate the probability of failing at each try until the (k-1)th try, and then calculate the probability of success on the kth try. Multiply these probabilities to get the required probability. Let Q(k) be the probability of opening the door at the kth try. \(Q(k) = \frac{n-1}{n} \cdot \frac{n-1}{n} \cdots \frac{n-1}{n} \cdot \frac{1}{n} \) Since there are (k-1) failures, the term \(\frac{n-1}{n}\) will be multiplied (k-1) times.

Step 3: Simplify the result

\(Q(k) = \left(\frac{n-1}{n}\right)^{k-1}\cdot\frac{1}{n} \) Therefore, the probability of opening the door on her kth try in this case is: \( \left(\frac{n-1}{n}\right)^{k-1}\cdot\frac{1}{n} \)

Key Concepts

Random Experiment

A random experiment is an action or process that results in one of several possible outcomes. These outcomes are determined by chance, meaning they cannot be precisely predicted before the experiment is conducted. In the context of probability theory, a random experiment forms the basis for analyzing probability events.
A simple example of a random experiment is flipping a coin. Here, you can either get "heads" or "tails," each with an equal likelihood. Similarly, the scenario of a woman trying different keys to open a door can be considered a random experiment. Each key she tries has a certain probability of being the right one, depending on various factors like whether she discards the keys after trying.

• If keys are discarded after each unsuccessful attempt, the number of possible outcomes decreases with each try, affecting the probability of success on each subsequent trial.
• If keys are not discarded, the probability remains unchanged for each attempt as all keys are still available to try.

Random experiments are crucial for calculating probabilities and understanding how likely a particular outcome is.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already happened. It provides a way to update our understanding of the likelihood of an event based on new information.
In mathematical terms, if you want to find out how likely event A is to happen, knowing that event B has already occurred, you use conditional probability. It's notably applied through the formula:\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]where \(P(A|B)\) is the probability of A given B, \(P(A \cap B)\) is the probability that both events A and B occur, and \(P(B)\) is the probability that event B occurs.
In relation to our key scenario:

• When keys are discarded, the probability changes as the woman progresses because she learns more with each try.
• When keys are not discarded, the condition is that the keys remain constant, so the conditional probability doesn't change as dramatically with each attempt.

These scenarios reflect the basics of conditional probability, where the probability of future attempts depends on what happened previously.

Probability Distribution

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes of a random experiment. It describes how probabilities are distributed over the events.
There are several types of probability distributions, like discrete and continuous distributions. Discrete distributions deal with countable outcomes, while continuous distributions relate to outcomes over a continuous range.
For the exercise at hand, the probability distribution function provides insights into the likelihood of each key-opening attempt being successful:

• **When keys are discarded**: The distribution is affected by the reduced number of keys. The probability function becomes \( \frac{n - (k - 1)}{n} \), highlighting that with each discarded key, the probability of selecting the correct key on the next try increases.
• **When keys are not discarded**: The function remains \( \left(\frac{n-1}{n}\right)^{k-1} \cdot \frac{1}{n} \), indicating the same chance in every trial irrespective of past attempts.

Understanding probability distributions helps in predicting outcomes and making data-driven decisions in uncertain scenarios.