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Magazine Chapter 3: Problem 45
Suppose that five people, including you and a friend, line up at random. Let the random variable \(X\) denote the number of people standing between you and your friend. What is \(p_{X}(k)\) ?
Short Answer
Expert verified
\[p_{X}(k)\] is 0.067 for \(k=0\), 0.15 for \(k=1\), 0.4 for \(k=2\), and 0.033 for \(k=3\).
Step by step solution
01
Definitions
Define the space of all possibilities. There are \(5!\) (factorial, which means multiplying 5*4*3*2*1) possible orderings of all five people. This is because 1st place can be occupied by any one of the 5 people, 2nd place can be occupied by any one of the remaining 4 people, and so on.
02
Cases for k
Now think of possible values for \(k\). The random variable \(X\) which represents the number of people between you and your friend, can take on values 0, 1, 2, 3. These are basically standing next to each other (k=0), one person between (k=1), two people between (k=2), and three people between (k=3). Therefore, we will calculate \(p_{X}(k)\) for each of these values of \(k\).
03
Calculating \(p_{X}(k)\) for each k
For \(p_{X}(k)\) where \(k=0\), it means you and your friend stand next to each other. You and your friend can stand on any of the 4 'places' in the line, and either you or your friend can be in front, which gives rise to two possibilities per place, so total \(4*2=8\) possibilities. Thus, \(p_{X}(0)=8/5!=0.067\).For \(p_{X}(k)\) where \(k=1\), it means one person stands between you and your friend. You, your friend, and the other person can be arranged in \(3!\) ways. The group of three can be in any of the 3 'places' in the line, so total \(3!*3=18\) possibilities. Thus, \(p_{X}(1)=18/5!=0.15\).For \(p_{X}(k)\) where \(k=2\), it means two people stand between you and your friend. You, your friend, and the other two can be arranged in \(4!\) ways. The group of four can be in any of the 2 'places' in the line, so total \(4!*2=48\) possibilities. Thus, \(p_{X}(2)=48/5!=0.4\).Lastly, for \(p_{X}(k)\) where \(k=3\), it means all three other people stand between you and your friend. You, your friend, and the other three can only be arranged in one way (you, 3 people, friend or friend, 3 people, you), so only 2 possibilities. Thus, \(p_{X}(3)=2/5!=0.033\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Random Variable
In probabilistic terms, a random variable is a variable that assumes numerical values related to the outcomes of a random phenomenon and it is essential for solving problems like our textbook example, where the lineup order of people is random. In this scenario, the variable we're referring to is denoted as \(X\), which represents the number of people between you and your friend. Random variables often serve as bridges connecting real-world random processes to mathematical frameworks, enabling us to use algebra and calculus to analyze probabilities and their distributions.
To visualize it, think of rolling a die. The outcome can be any number between 1 and 6, and each outcome corresponds to a random variable's value. Similarly, here \(X\) can take on several discrete values depending on how many people are interspersed between you and your friend.
To visualize it, think of rolling a die. The outcome can be any number between 1 and 6, and each outcome corresponds to a random variable's value. Similarly, here \(X\) can take on several discrete values depending on how many people are interspersed between you and your friend.
Factorial
The concept of factorial is a mathematical function denoted by an exclamation point (!) and is a key player when evaluating permutations and combinations. In this exercise, the notation \(5!\) means the product of all positive integers from 1 to 5. The calculation unfolds as \(5 \times 4 \times 3 \times 2 \times 1 = 120\), representing the total number of different ways five distinct items can be arranged, without repetition.
Understanding factorials is crucial in probability because it helps us determine the total number of possible outcomes, giving us a starting point from which to calculate probabilities. In our example problem, factorials help us calculate the total number of ways to arrange people in line.
Understanding factorials is crucial in probability because it helps us determine the total number of possible outcomes, giving us a starting point from which to calculate probabilities. In our example problem, factorials help us calculate the total number of ways to arrange people in line.
Permutations
Permutations allow us to count how many different ways we can arrange a set number of objects. They matter in our exercise because we are interested in the different orderings of people in the lineup. A key point in permutations is that order matters; the sequence in which people stand is important. The number of permutations of \(n\) distinct objects taken all at a time is given by \(n!\), which was the starting point in our example with five people.
When we look at our problem, we also consider permutations when determining how many ways you and your friend can be placed with a certain number of people between you. For instance, if there is one person between you, the sequencing of the three of you is given by a smaller factorial, \(3!\), as explained in the solution.
When we look at our problem, we also consider permutations when determining how many ways you and your friend can be placed with a certain number of people between you. For instance, if there is one person between you, the sequencing of the three of you is given by a smaller factorial, \(3!\), as explained in the solution.
Discrete Probability
Discrete probability involves scenarios where the outcome can only take on specific, distinct values, like flipping a coin which can only land heads or tails. In our exercise, the random variable \(X\) is an example of a discrete random variable, as it can only take on countable values: 0, 1, 2, or 3.
To compute the discrete probability \(p_X(k)\) of each possible value of \(X\), we divide the number of ways the specific event can happen by the total number of possible outcomes (which we figured out thanks to the concept of factorial). The solutions show this calculation for different values of \(k\) and exemplify how understanding probabilities requires not just a grasp of the variables themselves but also the context鈥攈ere, how people are positioned relative to one another in a lineup.
To compute the discrete probability \(p_X(k)\) of each possible value of \(X\), we divide the number of ways the specific event can happen by the total number of possible outcomes (which we figured out thanks to the concept of factorial). The solutions show this calculation for different values of \(k\) and exemplify how understanding probabilities requires not just a grasp of the variables themselves but also the context鈥攈ere, how people are positioned relative to one another in a lineup.
