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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

Suppose that in a certain metropolitan area, \(90 \%\) of all households have cable TV. Let \(x\) denote the number among four randomly selected households that have cable TV. Then \(x\) is a binomial random variable with \(n=4\) and \(p=0.9\). a. Calculate \(p(2)=P(x=2),\) and interpret this probability. b. Calculate \(p(4),\) the probability that all four selected households have cable TV. c. Calculate \(P(x \leq 3)\).

The accompanying data on \(x=\) student-teacher ratio is for a random sample of 20 high schools in Maine selected from a population of 85 high schools. The data are consistent with summary values for the state of Maine that appeared in an article in the Bangor Daily News (September \(22,2016,\) bangordailynews.com/2016/09/22/mainefocus/we-discovered-a-surprise-when-we- looked-deeper-into-our-survey-of-maine-principals/?ref=morelnmidcoast, retrieved May 2, 2017). The corresponding normal scores are also shown. $$ \begin{array}{|cc|} \hline \text { Student-Teacher Ratio }(x) & \text { Normal Score } \\ \hline 9.0 & -1.868 \\ 10.0 & -1.403 \\ 11.0 & -1.128 \\ 11.2 & -0.919 \\ 11.6 & -0.744 \\ 11.7 & -0.589 \\ 11.8 & -0.448 \\ 11.9 & -0.315 \\ 12.0 & -0.187 \\ 12.1 & -0.062 \\ 12.5 & 0.062 \\ 12.6 & 0.187 \\ 13.0 & 0.315 \\ 13.2 & 0.448 \\ 13.6 & 0.589 \\ 13.7 & 0.744 \\ 14.0 & 0.919 \\ 14.5 & 1.128 \\ 14.9 & 1.403 \\ 15.0 & 1.868 \\ \hline \end{array} $$ a. Construct a normal probability plot. b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the distribution of student-teacher ratios for high schools in Maine is approximately normal.

Suppose that \(25 \%\) of the fire alarms in a large city are false alarms. Let \(x\) denote the number of false alarms in a random sample of 100 alarms. Approximate the following probabilities: a. \(P(20 \leq x \leq 30)\) b. \(P(20

Suppose that \(20 \%\) of all homeowners in an earthquakeprone area of California are insured against earthquake damage. Four homeowners are selected at random. Define the random variable \(x\) as the number among the four who have earthquake insurance. a. Find the probability distribution of \(x\). (Hint: Let \(S\) denote a homeowner who has insurance and \(\mathrm{F}\) one who does not. Then one possible outcome is SFSS, with probability (0.2)(0.8)(0.2)(0.2) and associated \(x\) value of 3 . There are 15 other outcomes.) b. What is the most likely value of \(x ?\) c. What is the probability that at least two of the four selected homeowners have earthquake insurance?

A pizza shop sells pizzas in four different sizes. The 1000 most recent orders for a single pizza resulted in the following proportions for the various sizes: $$ \begin{array}{lcccc} \text { Size } & 12 \text { in. } & 14 \text { in. } & 16 \text { in. } & 18 \text { in. } \\ \text { Proportion } & 0.20 & 0.25 & 0.50 & 0.05 \end{array} $$ With \(x=\) the size of a pizza in a single-pizza order, the given table is an approximation to the population distribution of \(x\). a. Write a few sentences describing what you would expect to see for pizza sizes over a long sequence of single-pizza orders. b. What is the approximate value of \(P(x<16)\) ? c. What is the approximate value of \(P(x \leq 16)\) ?

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