Problem 80
Question
An elevator starts with \(m\) passengers and stops at \(n\) floors \((m \leq n)\). The probability that no two passengers alight at the same floor is (A) \(\frac{n_{P_{m}}}{m^{n}}\) (B) \(\frac{{ }^{n_{P}} P_{\mathrm{m}}}{n^{\mathrm{m}}}\) (C) \(\frac{n_{m_{m}}}{m^{\text {n }}}\) (D) \(\frac{n_{C_{m}}}{n^{\text {mu }}}\)
Step-by-Step Solution
Verified Answer
The answer is (B) \(\frac{{ }^{n_{P}} P_{\mathrm{m}}}{n^{\mathrm{m}}}\).
1Step 1: Understand the Problem
We have an elevator with \(m\) passengers and \(n\) floors. We need to find the probability that no passenger gets off on the same floor. This means we want each of the \(m\) passengers to alight on a different floor.
2Step 2: Determine the Total Possible Outcomes
For each passenger, there are \(n\) possible floors to choose from. Therefore, there are a total of \(n^m\) possible ways for \(m\) passengers to alight among \(n\) floors.
3Step 3: Determine the Favorable Outcomes
In order for no two passengers to alight on the same floor, we must select \(m\) different floors from \(n\) floors and then assign one passenger to each floor. The number of ways to choose \(m\) different floors from \(n\) available floors is given by the permutation \(_nP_m\).
4Step 4: Calculate the Probability
The probability is the ratio of favorable outcomes to the total outcomes:\[\text{Probability} = \frac{_nP_m}{n^m}\]
5Step 5: Compare Options
Compare the calculated probability to the given options.Option (B) is \(\frac{{ }^{n_{P}} P_{\mathrm{m}}}{n^{\mathrm{m}}}\) which matches exactly with the probability we calculated.
Key Concepts
PermutationsCounting PrinciplesProbability Calculations
Permutations
Permutations play a crucial role in problems where we need to arrange or order items. In our elevator problem, we deal with arranging passengers on floors. Here, we're not just picking any floor but assigning passengers to distinct floors, which is a classic permutation problem.
Permutations involve finding the number of ways to arrange a set of items. For our scenario, it means selecting and ordering exactly which passenger goes to which floor without any repetition. This concept distinguishes permutations from combinations, where order does not matter.
The notation used here is P_m, which denotes the number of permutations of selecting and arranging items. This is calculated using the formula:\[ _nP_m = \frac{n!}{(n-m)!} \]
This formula applies perfectly to our example where we are choosing and arranging floors for passengers, ensuring no two get off on the same floor.
Permutations involve finding the number of ways to arrange a set of items. For our scenario, it means selecting and ordering exactly which passenger goes to which floor without any repetition. This concept distinguishes permutations from combinations, where order does not matter.
The notation used here is P_m, which denotes the number of permutations of selecting and arranging items. This is calculated using the formula:\[ _nP_m = \frac{n!}{(n-m)!} \]
This formula applies perfectly to our example where we are choosing and arranging floors for passengers, ensuring no two get off on the same floor.
Counting Principles
Counting principles help break down complex problems into manageable steps by determining how many ways an event can occur. In the elevator scenario, our goal is to account for every possible arrangement of passengers getting off at different floors.
We start by considering each passenger and how many choices they have. The fundamental principle of counting provides a simple way to think about this: For each passenger (or event), multiply the number of possible choices available. If a passenger can choose any of the n floors, the total number of different combinations they can choose (without restriction) is given by multiplying for each passenger, resulting in \(n^m\), where \(m\) is the number of passengers.
When we apply conditions—like no two passengers alighting on the same floor—we further refine our count. This involves only considering the permutations, essentially eliminating arrangements where two passengers share a floor.
We start by considering each passenger and how many choices they have. The fundamental principle of counting provides a simple way to think about this: For each passenger (or event), multiply the number of possible choices available. If a passenger can choose any of the n floors, the total number of different combinations they can choose (without restriction) is given by multiplying for each passenger, resulting in \(n^m\), where \(m\) is the number of passengers.
When we apply conditions—like no two passengers alighting on the same floor—we further refine our count. This involves only considering the permutations, essentially eliminating arrangements where two passengers share a floor.
Probability Calculations
Probability is all about determining the likelihood of an event occurring out of all possible events. For the elevator problem, we're interested in finding the probability that each passenger gets off at a different floor.
To solve this, we start by calculating the total number of possible outcomes, which is \(n^m\), as each passenger has \(n\) floors to choose from independently. Next, we find the number of favorable outcomes, which is how often passengers can alight at different floors—calculated by permutations, P_m.
Probability is then the ratio of these favorable outcomes to the total possible outcomes:
\[\text{Probability} = \frac{_nP_m}{n^m} \]
This formula gives us a clear way to compare favorable configurations against all potential configurations, offering a practical method to calculate the probability the way it is done in option (B) in the original exercise.
To solve this, we start by calculating the total number of possible outcomes, which is \(n^m\), as each passenger has \(n\) floors to choose from independently. Next, we find the number of favorable outcomes, which is how often passengers can alight at different floors—calculated by permutations, P_m.
Probability is then the ratio of these favorable outcomes to the total possible outcomes:
\[\text{Probability} = \frac{_nP_m}{n^m} \]
This formula gives us a clear way to compare favorable configurations against all potential configurations, offering a practical method to calculate the probability the way it is done in option (B) in the original exercise.
Other exercises in this chapter
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