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There are \(n\) socks, 3 of which are red, in a drawer. What is the value of \(n\) if when 2 of the socks are chosen randomly, the probability that they are both red is \(\frac{1}{2} ?\)

Short Answer

Expert verified
The value of \(n\) is 4, meaning there are four socks in the drawer, 3 of which are red.

Step by step solution

01

Identify known values

We know that there are 3 red socks in the drawer and the probability of choosing two red socks is \(\frac{1}{2}\).
02

Express the probability as a fraction

Let's express the probability of choosing two red socks as a fraction of choosing 2 socks from the drawer in general. We can write this as: \[\frac{\text{Number of ways to choose 2 red socks}}{\text{Total number of ways to choose 2 socks}} = \frac{1}{2}\]
03

Calculate combinations

To calculate the number of ways to choose 2 red socks and the total number of ways to choose 2 socks, we will use combinations. The total number of ways to choose 2 socks from the drawer can be denoted as: \(C(n,2)\), which means we are choosing 2 socks out of n available socks. Similarly, the total number of ways to choose 2 red socks can be denoted as: \(C(3,2)\), which means we are choosing 2 socks out of 3 red socks. Therefore, we can rewrite the equation from Step 2 as: \[\frac{C(3,2)}{C(n,2)} = \frac{1}{2}\]
04

Apply the combination formula

The formula for combination can be defined as: \[C(n,r) = \frac{n!}{r!(n-r)!}\] Applying the combination formula for \(C(3,2)\) and \(C(n,2)\), we get: \[\frac{\frac{3!}{2!(3-2)!}}{\frac{n!}{2!(n-2)!}} = \frac{1}{2}\]
05

Simplify the equation

Let's simplify the equation by cancelling out the common terms: \[\frac{\frac{3!}{2!1!}}{\frac{n!}{2!(n-2)!}} = \frac{1}{2}\] \[\frac{3!}{1!} \times \frac{2!(n-2)!}{n!} = \frac{1}{2}\] Now, recall that the factorial of a number, say 3, is \(3! = 3 \times 2 \times 1\). So, 3! = 6 and 2! = 2: \[\frac{6}{1} \times \frac{2(n-2)!}{n!} = \frac{1}{2}\]
06

Solve for n

Now, we can multiply both sides by 2 and simplify further to solve for \(n\): \[6 \times \frac{(n-2)!}{n!} = 1\] \[(n-2)! = \frac{1}{6}n!\] Now we can try different values of n and see which one satisfies the equation. Given that there are only 3 red socks, we know that n must be greater than 3. Trying n=4, we see that it actually satisfies the equation: \[(4-2)! = \frac{1}{6}(4!)\] \[2! = \frac{1}{6}(24)\] \[2 = \frac{24}{6}\] \[2 = 4\] Since we have found a value for \(n\) which satisfies the given condition, we can conclude that:
07

State the final value of n

The value of \(n\) is 4, meaning there are four socks in the drawer, 3 of which are red.

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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.

When we talk about combinatory problems like the socks in the drawer, we're referring to the number of ways certain items can be selected or arranged. This particular problem involves combinations, a fundamental concept in combinatorics, where the order of selection does not matter. In other words, picking a red sock first and a blue sock second is the same as picking a blue sock first and then a red sock.

To understand this on a deeper level, imagine you have a variety of socks in the drawer, and you want to know how many unique pairs you can create. This is where combinatorial formulas, including those for permutations and combinations, come to play. In our example with the socks, since the order of drawing socks does not matter, we use the combination formula to find out the possible pairs.
Factorial Notation
When solving combinatorial problems, we often use factorial notation, denoted by an exclamation point (!). Factorial notation is a mathematical expression used to describe the product of an integer and all the positive integers below it. For example, saying '5 factorial' is the same as saying the product of 5 times 4 times 3 times 2 times 1, denoted as 5!.

However, by convention, 0! is always equal to 1. This is a fundamental rule in combinatorics that facilitates calculations, especially when working with the combination formulas, where 0! appears frequently.

In reference to our sock example, understanding how to calculate factorial is crucial. You'd calculate the factorial of the total number of socks (n!) and of smaller groups, like the number of red socks (3!), to figure out the different ways to pick the socks. Exploring large numbers quickly reveals how rapidly factorials can grow, which is a fun (and sometimes daunting) feature of this concept.
Basic Probability
Probability is all about measuring the chance of an event occurring, usually expressed as a fraction or percentage. Think of it as trying to predict the likelihood of pulling a red sock from a drawer without looking. To calculate basic probability, we need two main components: the number of favorable outcomes and the total number of possible outcomes.

Using our exercise, the event we care about is drawing two red socks. The number of favorable outcomes is the number of ways to choose 2 socks from the 3 red ones available. The total number of outcomes is how many different pairs of socks could be pulled from the drawer. So, the probability is a ratio of these two numbers.

With the concept of basic probability, students can tackle a wide array of real-life scenarios beyond theoretical exercises. From predicting weather events to making informed decisions in games of chance, understanding probability is a valuable life skill that goes hand in hand with combinatory logic and factorial notation.

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Most popular questions from this chapter

A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be (a) no complete pair; (b) exactly 1 complete pair?

An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly (a) all 5 problems; (b) at least 4 of the problems?

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In a state lottery, a player must choose 8 of the numbers from 1 to 40 . The lottery commission then performs an experiment that selects 8 of these 40 . numbers. Assuming that the choice of the lottery commission is equally likely to be any of the \(\left(\begin{array}{c}40 \\ 8\end{array}\right)\) combinations, what is the probability that a player has (a) all 8 of the numbers selected; (b) 7 of the numbers selected; (c) at least 6 of the numbers selected?

An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. These classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French. class, and 16 in the German class. There are 12 students that are in both Spanish and French, 4 that are in both Spanish and German, and 6 that are in both French and German. In addition, there are 2 students taking all 3 classes. (a) If a student is chosen randomly, what is the probability that he or she is not in any of these classes? (b) If a student is chosen randomly, what is the probability that he or she is taking exactly one language class? (c)) If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

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