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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

The paper referenced in the previous exercise also included data on left atrial diameter for children who were considered overweight. For these children, left atrial diameter was approximately normally distributed with a mean of \(28 \mathrm{~mm}\) and a standard deviation of \(4.7 \mathrm{~mm}\). a. Approximately what proportion of overweight children have left atrial diameters less than \(25 \mathrm{~mm}\) ? b. Approximately what proportion of overweight children have left atrial diameters greater than \(32 \mathrm{~mm} ?\) c. Approximately what proportion of overweight children have left atrial diameters between 25 and \(30 \mathrm{~mm}\) ? d. What proportion of overweight children have left atrial diameters greater than the mean for healthy children?

An automobile manufacturer is interested in the fuel efficiency of a proposed new car design. Six nonprofessional drivers were selected, and each one drove a prototype of the new car from Phoenix to Los Angeles. The resulting fuel efficiencies \((x,\) in miles per gallon \()\) are: $$ \begin{array}{llllll} 27.2 & 29.3 & 31.2 & 28.4 & 30.3 & 29.6 \end{array} $$ The normal scores for a sample of size 6 are $$ \begin{array}{llllll} -1.282 & -0.643 & -0.202 & 0.202 & 0.643 & 1.282 \end{array} $$ a. Construct a normal probability plot for the fuel efficiency data. Does the plot look linear? 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 fuel efficiency distribution is approximately normal.

Suppose that the amount of time spent by a statistical consultant with a client at their first meeting is a random variable that has a normal distribution with a mean value of 60 minutes and a standard deviation of 10 minutes. a. What is the probability that more than 45 minutes is spent at the first meeting? b. What amount of time is exceeded by only \(10 \%\) of all clients at a first meeting?

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

An appliance dealer sells three different models of freezers having 13.5,15.9 , and 19.1 cubic feet of storage space. Consider the random variable \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: \(x\) \(\begin{array}{lll}13.5 & 15.9 & 19.1\end{array}\) \(p(x)\) \(\begin{array}{lll}0.2 & 0.5 & 0.3\end{array}\) a. Calculate the mean and standard deviation of \(x\). (Hint: See Example \(6.15 .\) ) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

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