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Use counting principles from Section 10.4 to calculate the number of outcomes. A group of friends, five girls and five boys, wants to go to the movies on Friday night. The friends select, at random, two of their group to go to the ticket office to purchase the tickets. What is the probability that the two selected are both boys?

Short Answer

Expert verified
The probability that the two selected to go to the ticket office are both boys is 0.222.

Step by step solution

01

Calculate Total Outcomes

Calculate the total number of outcomes when 2 people are selected from a group of 10 friends. We use the combination formula \(C(n, r) = n! / [r!(n-r)!]\), where \(n\) is the total number of items, \(r\) is the items to choose, and \(!\) denotes factorial. Here, \(n\) is 10 (total friends) and \(r\) is 2 (friends to select). So, \(C(10, 2) = 10! / [2!(10-2)!] = 45.\)
02

Calculate Favorable Outcomes

Calculate the number of favorable outcomes, which is selecting 2 boys from 5. We use the same formula, but here, \(n\) is 5 (total boys) and \(r\) is 2 (boys to select). So, \(C(5, 2) = 5! / [2!(5-2)!] = 10.\)
03

Calculate Probability

The probability is the ratio of the number of favorable outcomes to the total number of outcomes which is given by the formula \(P = Favorable Outcomes / Total Outcomes\). Thus, \(P = 10 / 45 = 0.222.\)

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Key Concepts

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

Counting Principles
Understanding how to calculate different scenarios in probability often starts with the counting principles. These are fundamental concepts that establish the groundwork for determining the number of ways certain events can occur.

Two main counting principles are pertinent to many probability exercises: the multiplication rule and the addition rule. The multiplication rule states that if you want to find the number of outcomes for a sequence of events, you multiply the number of ways each event can occur. However, in our movie ticket example, the order of selection doesn't matter, so we don't use the multiplication rule directly.

The addition rule is applied when you calculate the number of outcomes when there are several different possible events, but again, for the movie ticket problem, it's about combinations of people rather than independent events that add up.

In summary, while these counting principles are the backbone of understanding more complex probability problems, for questions about combinations of a group, as in the movie ticket example, we delve further into combinations and factorials to calculate our probabilities accurately.
Combinations
Combinations are a key concept when dealing with problems where the order of selection isn't important. In our textbook scenario, we're looking at how to select 2 friends from a group of 10 without caring about the order they're chosen in - this is a classic use of combinations.

To understand combinations, one should grasp the idea of 'n choose r' — written as \( C(n, r) \). This represents the different ways you can choose \( r \) items from \( n \) without regard to the order. The formula for this, \( C(n, r) = \frac{n!}{r!(n-r)!} \), incorporates factorials (which we'll discuss in the next section). Here's a practical use of combinations in our problem:
  • The total number of ways to pick 2 friends from 10 is \( C(10, 2) = 45 \).
  • To find the probability of selecting 2 boys from a group of 5, we calculate \( C(5, 2) = 10 \).
These calculations allow us to solve our exercise’s question by providing a structured way to count without listing every possible group or pair, which becomes unfeasible with larger numbers.
Factorials
The factorial is a mathematical operation that is pivotal to computing combinations and, by extension, to solving many probability problems. Represented by an exclamation mark \( ! \), a factorial is the product of all positive integers less than or equal to a certain number. For instance, the factorial of 5 or \( 5! \) is calculated as \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow extremely fast, making manual calculations cumbersome for large numbers. That's why for larger numbers, it's always better to use a calculator or software.

Factorials are not only used when calculating the number of permutations (where the order is important) but also for combinations. The formula for combinations, as seen in the textbook exercise, includes factorials in both the numerator and denominator which helps reduce the expression to a more manageable number, as we did when calculating the number of ways to pick 2 boys from a group of 5. This simplification is crucial for smoothly finding the solution without getting lost in the multiplication of large numbers.

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Most popular questions from this chapter

This set of exercises will draw on the ideas presented in this section and your general math background. The first term of an arithmetic sequence is \(4 .\) The sum of the first three terms of the sequence is \(24 .\) Use summation notation to express the sum of the first eight terms of this sequence, and use an appropriate formula to find the sum.

In this set of exercises, you will use sequences to study real-world problems. Investment An income-producing investment valued at \(\$ 3000\) pays interest at an annual rate of \(4.5 \% .\) Assume that the interest is taken out as income and therefore is not compounded. (a) Make a table in which you list the initial investment along with the total value of the investment-related assets (initial investment plus total interest earned) at the end of each of the first 4 years. (b) What is the total value of the investment-related assets after \(n\) years?

Induction is not the only method of proving that a statement is true. Exercises \(26-29\) suggest alternate methods for proving statements. Prove that \(1+4+4^{2}+\cdots+4^{n-1}=\frac{4^{n}-1}{3}\) by using the formula for the sum of terms of a geometric sequence.

This set of exercises will draw on the ideas presented in this section and your general math background. What are the terms of the sequence generated by the expression \(a_{n}=a_{0}+n d, d=0 ?\)

Write out all the different four-digit numbers possible using the numbers \(1,1,2,3 .\) Why is your number of possibilities not equal to 4!?

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