/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 A man has \(n\) keys on a key ri... [FREE SOLUTION] | 91Ó°ÊÓ

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A man has \(n\) keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he returns home only to find himself unable to distinguish one key from another. Resourceful, he works out a fiendishly clever plan: He will choose a key at random and try it. If it fails to open the door, he will discard it and choose at random one of the remaining \(n-1\) keys, and so on. Clearly, the probability that he gains entrance with the first key he selects is \(1 / n\). Show that the probability the door opens with the third key he tries is also \(1 / n\). (Hint: What has to happen before he even gets to the third key?)

Short Answer

Expert verified
The probability that the door opens with the third key is \(1 / n\).

Step by step solution

01

Calculate the probability of the first key not opening the door

First key not being the right key would be \((n-1) / n\).
02

Calculate the probability of the second key also not opening the door

If the first key did not work, we are left with \(n-1\) keys. The probability of the second key also not being correct would be \((n-2) / (n-1)\).
03

Calculate the probability of the third key opening the door

If the first two keys did not work, we are left with \(n-2\) keys. Since one of these keys is the one that can open the door, the probability that the third key is the right one is \(1 / (n-2)\).
04

Calculate the total probability

The total probability of the door opening at the third key is the product of these three probabilities, i.e. \((n-1) / n * (n-2) / (n-1) * 1 / (n-2) = 1 / n\). Please remember that the man discards unsuccessful keys which alters the number of total possibilities at each step, so the probabilities are not independent and can't be calculated separately.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Understanding probability theory is essential for making sense of uncertain events. Consider a man who has a key ring with an unknown number of keys, but only one that opens his door. When he randomly selects a key to try to open the door, the basic tenet of probability theory comes into play: the likeliness of an event happening. The probability of selecting the correct key on the first try is simply one out of the total number of keys, calculated as \(\frac{1}{n}\).

However, probability theory also involves understanding what happens when an event does not occur. In our key example, if the first key fails, the man eliminates that key from the subsequent draw—a process known as conditional probability. The chance of success on the second try changes because the total number of keys has decreased. These dynamic changes affect the calculated likelihoods at each step of the process, illustrating how probability theory can model complex, dependent events.
Combinatorics
Combinatorics is the branch of mathematics dealing with combinations and arrangements of objects. In the context of our key example, combinatorics helps in analyzing the different ways to select keys. Initially, the man has \(n\) keys, and there are \(n\) possible choices for the first key to try. Once he discards one key, there are \(n-1\) options left, and so on.

Although this may seem like a simple decrement, the foundational principles of combinatorics are at work. Counting principles, such as the rule of product, highlight how the total number of outcomes changes as the situation progresses. By the third key, the man's choices have been reduced, yet combinatorically, he still has a certain number of scenarios to consider, with each scenario holding equal weight, thus ensuring that the probability of the third key being correct remains \(\frac{1}{n}\).
Statistical Reasoning
Statistical reasoning allows for making sense of data and chance in a logical manner. In our scenario, statistical reasoning is used to understand the man's likelihood of success across several trials with his keys. It's understandable to think that his chances improve with each failed attempt, but statistics provide a counterintuitive insight: despite previous failures, the probability of successfully opening the door with the next key remains constant at \(\frac{1}{n}\).

This is because each key attempt is a separate 'experiment' with its own success and failure rates, based on the remaining keys. The process highlights an important aspect of statistical reasoning—the interplay between independent and dependent events and how to properly assess probabilities within that framework. The idea of eliminating unsuccessful options and therefore changing the sample space is a pivotal concept that statistical reasoning helps to clarify.

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Most popular questions from this chapter

Let \(A\) and \(B\) be two events defined on a sample space \(S\) such that \(P\left(A \cap B^{C}\right)=0.1, P\left(A^{C} \cap B\right)=0.3\), and \(P\left((A \cup B)^{C}\right)=0.2\). Find the probability that at least one of the two events occurs given that at most one occurs.

School board officials are debating whether to require all high school seniors to take a proficiency exam before graduating. A student passing all three parts (mathematics, language skills, and general knowledge) would be awarded a diploma; otherwise, he or she would receive only a certificate of attendance. A practice test given to this year's ninety-five hundred seniors resulted in the following numbers of failures: $$ \begin{array}{lc} \hline \text { Subject Area } & \text { Number of Students Failing } \\ \hline \text { Mathematics } & 3325 \\ \text { Language skills } & 1900 \\ \text { Ceneral knowledge } & 1425 \\ \hline \end{array} $$ If "Student fails mathematics," "Student fails language skills," and "Student fails general knowledge" are independent events, what proportion of next year's seniors can be expected to fail to qualify for a diploma? Does independence seem a reasonable assumption in this situation?

An automobile insurance company has compiled the information summarized below on its policy-holders. Suppose someone calls to file a claim. To which age group does he or she most likely belong? $$ \begin{array}{lcc} \hline \text { Age Group } & \begin{array}{c} \% \text { of } \\ \text { Policyholders } \end{array} & \begin{array}{c} \% \text { Involved in } \\ \text { Accidents Last Year } \end{array} \\ \hline \text { Young }(<30) & 20 & 35 \\ \text { Middle-aged }(30-64) & 50 & 15 \\ \text { Elderly }(65+) & 30 & 25 \\ \hline \end{array} $$

Stanley's statistics seminar is graded on a Pass/Fail basis. At the end of the semester each student is given the option of taking either a two-question exam (Final A) or a three-question exam (Final B). To pass the course, students must answer at least one question correctly on whichever exam they choose. The professor estimates that a typical student has a \(45 \%\) chance of correctly answering each of the two questions on Final A and a \(30 \%\) chance of correctly answering each of the three questions on Final B. Which exam should Stanley choose? Answer the question two different ways.

In a roll of a pair of fair dice (one red and one green), let \(A\) be the event of an odd number on the red die, let \(B\) be the event of an odd number on the green die, and let \(C\) be the event that the sum is odd. Show that any pair of these events is independent but that \(A, B\), and \(C\) are not mutually independent.

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