/*! 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 102 A box contains five slips of pap... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(5 .\) An outcome of the experiment will consist of two of these numbers.)

Short Answer

Expert verified
The probability distribution of the random variable \(w\) is: \(P(w=\$1) = \frac{3}{10}\) \(P(w=\$10) = \frac{4}{10}\) \(P(w=\$25) = \frac{3}{10}\).

Step by step solution

01

Calculating individual probabilities

There are 10 different pairs and, since each pair is equally likely, the probability of selecting any specific pair is 1/10. Now we can group the outcomes by the amount awarded: 1. Three pairs result in \(\$1\) being the largest amount: \((1,2), (1,3),\) and \((2,3)\). 2. Four pairs result in \(\$10\) being the largest amount: \((1,4), (2,4), (3,4),\) and \((4,5)\). 3. Three pairs result in \(\$25\) being the largest amount: \((1,5), (2,5),\) and \((3,5)\).
02

Calculating probabilities for each group

We will now calculate the probability of each amount being awarded: 1. Probability of \(w=\$1\): There are 3 pairs resulting in \(\$1\), so the probability is \(3/10\). 2. Probability of \(w=\$10\): There are 4 pairs resulting in \(\$10\), so the probability is \(4/10\). 3. Probability of \(w=\$25\): There are 3 pairs resulting in \(\$25\), so the probability is \(3/10\). The probability distribution of the random variable \(w\) is: \(P(w=\$1) = \frac{3}{10}\) \(P(w=\$10) = \frac{4}{10}\) \(P(w=\$25) = \frac{3}{10}\).

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability theory that assigns numerical values to the outcomes of random experiments. Imagine flipping a coin; the result could be heads or tails. If we assign a value of 1 to heads and 0 to tails, we have defined a random variable for our coin flip experiment.

In the provided exercise, the random variable, denoted as \(w\), is defined as the higher dollar amount selected from two slips of paper. The distinct possible values for \(w\) are \(1, \)10, and $25, which are derived from the combinations of slips that can be chosen. Understanding random variables is crucial as they are the foundation for calculating probabilities and distributions, and they can be either discrete, like in this example, or continuous.
Probability Calculation
Probability calculation is the process of determining the likelihood of a particular outcome. In the context of our example, each of the slips of paper has an equal chance of being selected, making this a uniform probability model. To find the probability of each potential monetary outcome from the box, we simply count the favorable outcomes and divide by the total number of possible outcomes.

For instance, there are 10 unique pairs that can be drawn, and each pair has a 1/10 chance of being selected. When we calculate the groups based on the largest value of each pair, as demonstrated in the exercise, we arrive at the probabilities for each amount. Determining these probabilities requires an understanding of basic combinatorics to ensure all possible outcomes are accounted for.
Combinatorics
Combinatorics is the area of mathematics dealing with counting, combination, and permutation of sets. It plays a key role in probability calculation by helping to determine the number of ways certain events can occur. To understand the probability distribution in our exercise, we need to use combinatorics to count all the possible pairs that can be drawn from the slips.

Since the slips can be considered as distinct objects, we can use combinations to find out the total number of unique pairs. The concept of combinations tells us that the number of ways to choose 2 items from a set of 5, without regard to order, is given by the formula \(\frac{5!}{2!(5-2)!}\), which simplifies to 10. This combinatorial reasoning helps us ensure that our probability calculations account for all possible outcomes without duplication, leading to a more accurate probability distribution.

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Most popular questions from this chapter

Twenty-five percent of the customers of a grocery store use an express checkout. Consider five randomly selected customers, and let \(x\) denote the number among the five who use the express checkout. a. Calculate \(p(2),\) that is, \(P(x=2)\). b. Calculate \(P(x \leq 1)\). c. Calculate \(P(x \geq 2)\). (Hint: Make use of your answer to Part (b).) d. Calculate \(P(x \neq 2)\).

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6.9 Two six-sided dice, one red and one white, will be rolled. List the possible values for each of the following random variables. a. \(x=\) sum of the two numbers showing b. \(y=\) difference between the number on the red die and the number on the white die (red - white) c. \(w=\) largest number showing

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