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From a box containing 4 dimes and 2 nickels, 3 coins are selected at random without replacement. Find the probability distribution for the total \(T\) of the 3 coins. Express the probability distribution graphically as a probability histogram.

Short Answer

Expert verified
The probability distribution is as follows: \(P(T=15) = \frac{{\text{{Comb}}(2,3)}}{{\text{{Comb}}(6,3)}}\), \(P(T=20) = \frac{{\text{{Comb}}(2,2) \times \text{{Comb}}(4,1)}}{{\text{{Comb}}(6,3)}}\), \(P(T=25) = \frac{{\text{{Comb}}(2,1) \times \text{{Comb}}(4,2)}}{{\text{{Comb}}(6,3)}}\), \(P(T=30) = \frac{{\text{{Comb}}(4,3)}}{{\text{{Comb}}(6,3)}}\). These results can be plotted into a histogram for a visual representation of the probability distribution.

Step by step solution

01

Identify the possible totals

First, one needs to identify the different possible outcomes i.e., sum of the coins in cents. Since there are dimes and nickels in the box, and 3 coins are selected, the possible totals could be 15 (three nickels), 20 (two nickels and one dime), 25 (one nickel and two dimes) and 30 (three dimes) cents. Thus, \(T\) could be either of 15, 20, 25 and 30.
02

Calculate combinations for each possible total

Calculate the combinations of coins that lead to each total. In the case of total value \(T=15\) cents, only combination is three nickels. For \(T=20\) cents, combination is 2 nickels and 1 dime. For \(T=25\) cents, combination is 1 nickel and 2 dimes. Finally, for \(T=30\) cents, the combination is 3 dimes.
03

Calculate the probabilities

Next, calculate the probability for each possible total. The probability that 3 coins sum up to a total \(T\) is equal to the number of combinations resulting to \(T\) divided by the total number of combinations of picking 3 coins without replacement, this can be calculated using combination formula. Here the probability of \(T=15\) cents, \(P(T=15) = \frac{{\text{{Comb}}(2,3)}}{{\text{{Comb}}(6,3)}}\), \(T=20\) cents, \(P(T=20) = \frac{{\text{{Comb}}(2,2) \times \text{{Comb}}(4,1)}}{{\text{{Comb}}(6,3)}}\), \(T=25\) cents, \(P(T=25) = \frac{{\text{{Comb}}(2,1) \times \text{{Comb}}(4,2)}}{{\text{{Comb}}(6,3)}}\), \(T=30\) cents, \(P(T=30) = \frac{{\text{{Comb}}(4,3)}}{{\text{{Comb}}(6,3)}}\) where \(\text{{Comb}}(n,k)\) stands for the number of ways to choose \(k\) elements from \(n\) elements
04

Express the probability distribution graphically

Finally, the probability distribution could be expressed graphically as a histogram. The values of \(T\) (15, 20, 25, 30) are represented on the x-axis, whereas the corresponding probabilities calculated in step 3 will be represented on the y-axis.

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Key Concepts

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

Combinatorics
Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is closely related to many other areas of mathematics and has many applications ranging from logic to statistical physics, from evolutionary biology to computer science, etc.

In the context of our exercise, combinatorics helps us to determine the number of possible outcomes when selecting coins from a box. The most fundamental principle of counting in combinatorics is the rule of product, which states that if an event can occur in 'm' ways and is followed by another event that can occur in 'n' ways, then the total number of ways the two events can occur is 'm x n'. In situations where order doesn't matter, we use combinations. A combination is a selection of items from a collection, such that the order of selection does not matter.

For our problem, to find out how many different ways we can pick 3 coins from a box of 6 coins, we use the combination formula, which is represented as \( \text{{Comb}}(n,k) \) where 'n' is the total number of items to choose from, 'k' is the number of items to pick, and \( \text{{Comb}}(n,k) = \frac{{n!}}{{k! \times (n - k)!}} \). Understanding combinatorics is crucial for solving problems related to probability, as seen in the step-by-step solution provided.
Probability Histogram
A probability histogram is a graphical representation of a probability distribution. It shows the probabilities of different outcomes of a discrete random variable as bars. The height of each bar corresponds to the probability of each outcome, and the total area of all the bars is equal to 1 (or 100%), as probabilities must sum up to 1.

In our exercise, the probability histogram would illustrate the probabilities of achieving a total of 15, 20, 25, or 30 cents with the selected coins. To construct the histogram, we place the different totals along the x-axis and their associated probabilities on the y-axis. Each outcome is represented by a bar rising to the height of its probability. Creating a histogram not only provides a visual representation of the probabilities but also makes it easier to understand the distribution of outcomes at a glance.

For example, if the probability of getting a total of 15 cents is 0.1, then the bar representing 15 cents would rise to 0.1 on the y-axis. The probability histogram is an excellent tool for visual learners and can assist all students in grasping the distribution of probabilities more intuitively.
Discrete Random Variable
A discrete random variable is a type of random variable that can take on a countable number of distinct outcomes. These outcomes are typically whole numbers, like the score on a dice roll or the number of books you might read in a month. Discrete random variables are the cornerstone of probability theory and are contrasted with continuous random variables, which can take on a continuum of values, like heights or weights.

In our exercise, the total value of the coins selected, denoted by \(T\), is a discrete random variable because it can only take on a certain number of values (15, 20, 25, or 30 cents). The probability distribution of a discrete random variable lists the probabilities associated with each of its possible values. It's crucial to understanding that the probability distribution must satisfy two conditions: the sum of the probabilities must be 1, and the probability of each outcome must be between 0 and 1 inclusive.

Grasping the concept of a discrete random variable is essential when dealing with problems like our coin selection exercise. It frames the problem within the realm of probability and allows us to use mathematical tools to calculate outcomes and their likelihoods, providing a clear pathway to solutions in numerous real-world scenarios.

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Most popular questions from this chapter

Each rear tire on an experimental airplane is supposed to be filled to a pressure of 40 pound per square inch (psi). Let \(X\) denote the actual air pressure for the right tire and \(Y\) denote the actual air pressure for the left tire. Suppose that \(X\) and \(Y\) are random variables with the joint density $$ f(x, y)=\left\\{\begin{array}{ll} k\left(x^{2}+y^{2}\right), & 30 \leq x<50; \\ & 30 \leq y<50; \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Find \(k\). (b) Find \(\mathrm{P}(30 \leq X \leq 40\) and \(40 \leq Y<50)\) (c) Find the probability that both tires are underfilled.

A cereal manufacturer is aware that the weight of the product in the box varies slightly from box to box. In fact, considerable historical data has allowed the determination of the density function that describes the probability structure for the weight (in ounces). In fact, letting \(X\) be the random variable weight, in ounces, the density function can be described as $$ f(x)=\left\\{\begin{array}{ll} \frac{2}{5}, & 23.75 \leq x \leq 26.25, \\ 0, & \text { elsewhere.} \end{array}\right. $$ (a) Verify that this is a valid density function. (b) Determine the probability that the weight is smaller than 24 ounces. (c) The company desires that the weight exceeding 26 ounces is an extremely rare occurrence. What is the probability that this "rare occurrence" does actually occur?

The total number of hours, measured in units of 100 hours, that a family runs a vacuum cleaner over a period of one year is a continuous random variable \(X\) that has the density function $$ f(x)=\left\\{\begin{array}{ll} x_{1} & 0 < x< 1, \\ 2-x, & 1 \leq x <2, \\ 0, & \text { elsewhere } \end{array}\right. $$ Find the probability that over a period of one year, a family runs their vacuum cleaner (a) less than 120 hours; (b) between 50 and 100 hours,

A continuous random variable \(X\) that can assume values between \(x=2\) and \(x=5\) has a density function given by \(f(x)=2(1+x) / 27\). Find (a) \(P(X<4) ;\) (b) \(\mathrm{P}(3 \leq X<4).\)

The probability distribution of \(A\), the number of imperfections per 10 meters of a synthetic fabric in continuous rolls of uniform width is given by $$ \begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 0.41 & 0.37 & 0.16 & 0.05 & 0.01 \end{array} $$ Construct the cumulative distribution function of \(X .\)

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