Question

\A box contains five slips of paper, marked $$\$ 1$$, $$\$ 1$$, $$\$ 1$$, $$\$ 10$$, and $$\$ 25 .$$ The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable \(w\) by \(w=\) amount awarded. Determine the probability distribution of \(w\). (Hint: Think of the slips as numbered \(1,2,3,4\), and 5 , so that an outcome of the experiment consists of two of these numbers.)

Short answer

The probability distribution of \(w\) is: \(P(w = 1) = 0.3\), \(P(w = 10) = 0.3\), \(P(w = 25) = 0.4\).

Step-by-step solution

Identify all possible combinations

The task is to select two slips of paper from five slips. Arrange slips in increasing order: \(1,1,1,10,25\). Assume that these correspond to numbers 1 to 5. We can see that it is possible to draw the slips in pairs: \( (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)\).

Calculate the maximum prize for each combination

Assign the value of each pair as such: \((1,2) - 1,\(1,3) - 1,\(1,4) - 10,\(1,5) - 25,\(2,3) - 1,\(2,4) - 10,\(2,5) - 25,\(3,4) - 10,\(3,5) - 25,\(4,5) - 25\). There are three possibilities of values \(w\) can take, 1, 10, and 25.

Calculate the probability for each case

Now, count the number of times each value occurs. - \(1\) occurs \(3\) times. - \(10\) occurs \(3\) times. - \(25\) occurs \(4\) times. There are a total of \(10\) possible outcomes so the probability for each value of \(w\) is calculated by dividing the number of times it occurs by \(10\). Hence, the probability distribution of \(w\) is - \(P(w = 1) = 3/10\), - \(P(w = 10) = 3/10\), - \(P(w = 25) = 4/10\).

Key concepts

Random Variable

A random variable is a numerical quantity that is assigned to each outcome of a random phenomenon. In our exercise, the random variable is denoted as \(w\), which represents the amount awarded to the winner of the contest based on the slips they select. The role of random variables in probability is crucial because they help us to quantify what we can expect to happen in uncertain situations. For instance, \(w\) in our case can take on any value that the contest rules allow, depending on which two slips are drawn.

When defining a random variable, we need to specify what values it can take and how to associate these values with the outcomes of our random process. For the contest example, there are only three values that \(w\) can assume, which are determined by the highest denomination of the two drawn slips: \$1, \$10, or \$25.

Probability Calculation

To perform a probability calculation, we need to understand how to evaluate the chance of a particular outcome occurring. The basic formula is the number of favorable outcomes divided by the total number of equally possible outcomes. In our exercise, the calculation involves determining the likelihood of each amount being the maximum one drawn.

Let's consider the exercise improvement advice, which suggests being detailed and selecting the most important topics. The probability of obtaining the maximum value from the slips is found by counting how many combinations yield each maximum value, then dividing by the total number of combinations possible. Here, the probability that \(w\) equals \$1, \$10, or \$25 is calculated by noting that \$1 occurs in three out of ten combinations, as does \$10, while \$25 appears in four combinations. Therefore, the probability distribution of \(w\) provides us with a clear picture of our chances for each potential amount awarded in the contest.

Statistical Combinations

The concept of statistical combinations tells us about the different ways we can combine a certain number of objects without regard to the order. It plays a central role in our understanding of probability because it helps us count how many possible outcomes there are. In this exercise, the number of combinations of two slips from five is calculated without considering the order in which they are drawn.

To clarify, drawing slip 1 then slip 2 is the same as drawing slip 2 then slip 1, since the order does not matter when determining the highest value. There are 10 unique combinations when drawing two slips out of five. Understanding how to enumerate these combinations is critical because it forms the denominator in the calculation of probabilities for each possible value of our random variable \(w\). It is the foundation on which we can build a probability distribution. This concept is a great tool for determining all possible outcomes, specifically in situations where different combinations lead to different results, as is the case in our exercise.