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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

Expert verified
The probability distribution of the amount awarded is: w(\$1)=0.3, w(\$10)=0.3, w(\$25)=0.4.

Step by step solution

01

Determine Possible Outcomes

Since two slips are drawn out of the five, the total possible outcomes can be calculated by using the combination formula. So there are \({{5}\choose{2}} = 10\) possible outcomes.
02

Determine Individual Outcomes

Next we determine the winning amount for each outcome. The winning amount 'w' is the larger value among the two slips selected. Thus, there can be three possible winning amounts: \$1, \$10 and \$25.
03

Calculate Probability for \$1

The winning amount will be \$1 only when both selected slips are \$1. As there are three \$1 slips out of five, the total outcomes for \$1 winning amount is \({{3}\choose{2}} = 3\). So, the probability of getting \$1 is 3/10.
04

Calculate Probability for \$10

The winning amount will be \$10 when either both slips are \$10 (which is not possible as there is only one \$10 slip) or when one slip is \$10 and other is \$1. There are 3 possible outcomes for the latter case, thus the probability of getting \$10 is 3/10.
05

Calculate Probability for \$25

The winning amount will be \$25 when either both slips are \$25 (which is not possible as there is only one \$25 slip) or when one slip is \$25 and the other is any other slip. There are 4 possible outcomes for the latter case, so probability of getting \$25 is 4/10=0.4.
06

Prepare the Probability Distribution

Now that the probabilities for all winning amount are known, the random variable w's distribution can be written as follows, w(\$1)=0.3, w(\$10)=0.3, w(\$25)=0.4.

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Key Concepts

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

Random Variable
In probability and statistics, a random variable is a numerical outcome of a random phenomenon. It associates each outcome of a chance experiment with a real number. In the exercise, the random variable is denoted by \( w \), representing the amount awarded from the contest. Imagine taking two slips from a box at random. The number each slip holds forms a part of this random variable. The primary role of \( w \) is to map these random outcomes (numbers on slips) to award amounts that one might receive, such as \(1, \)10, or $25.

  • The main point of a random variable is to simplify and summarize random outcomes numerically.
  • They are usually described using probability distributions to showcase the likelihood of various outcomes.
This makes it easier to think about and calculate probabilities associated with experiments, such as the chances of winning a particular amount in the contest.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, arranging, and finding combinations. When organizing random draws or selections, combinatorics helps calculate possible outcomes.In the given exercise, we used combinatorics to determine how many different pairs of slips can be selected from five.

We achieved this by using the combination formula, found with the symbol \( \binom{n}{k} \). Here, \( \binom{5}{2} \) equals 10, showing there are 10 different combinations of drawing two slips from five. Each combination represents a possible outcome, from which we derive the probabilities of winning different amounts.
Probability Calculation
Probability calculation is the process of determining the likelihood of different outcomes occurring. In the exercise, probabilities are calculated for deriving a particular reward \( w \). We start by observing the favorable outcomes for each potential award, such as \(1, \)10, and \(25.

For example, the probability of winning \)1 can be calculated by dividing the number of ways to draw two \(1 slips (3 ways) by the total possible outcomes (10 combinations). Thus, the probability is \( \frac{3}{10} \). Similarly, for \)10 and $25, we find probabilities of \( \frac{3}{10} \) and \( \frac{4}{10} \) respectively.
  • Probabilities must sum up to 1 when considering all possible outcomes.
  • They allow us to predict the likelihood of events in uncertain scenarios such as games or draws.
Discrete Probability
Discrete probability refers to a probability distribution where the random variable can take on a countable number of distinct values. In our exercise, \( w \) is a discrete random variable, as it can only take on three distinct values: \(1, \)10, and \(25. The probability distribution of \( w \) specifies the probability associated with each of these outcomes.

Discrete probability distributions are crucial for understanding risks and chances in games of chance, financial projections, and more. In this case, the distribution was described as:
  • \( w(\\)1) = 0.3 \)
  • \( w(\\(10) = 0.3 \)
  • \( w(\\)25) = 0.4 \)
This means that, out of all the possible outcomes, there's a 30% chance of winning \(1, a 30% chance of winning \)10, and a 40% chance of winning $25.

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Most popular questions from this chapter

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