Problem 19 - Algebra 2 and Trigonometry

Question

A variety box of instant oatmeal contains 10 plain, 6 maple, and 4 apple- cinnamon flavored packets. Ernestine reaches in and takes 3 packets without looking. Find each probability: $$ \begin{array}{ll}{\text { a. } P(2 \text { plain })} & {\text { b. } P(1 \text { maple, } 1 \text { apple-cinnamon })} \\ {\text { c } P(2 \text { plain, } 1 \text { maple })} & {\text { d. } P(1 \text { of each flavor) }}\end{array} $$

Step-by-Step Solution

Answer

a. \(\frac{15}{38}\), b. \(\frac{18}{95}\), c. \(\frac{9}{38}\), d. \(\frac{4}{19}\).

Step 1: Total number of ways to choose 3 packets

The variety box contains 20 total packets (10 plain, 6 maple, 4 apple-cinnamon). To find the total number of ways to choose 3 packets from the 20, we use the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items and \( r \) is the number of items to choose. Calculate \( \binom{20}{3} \): \[ \binom{20}{3} = \frac{20!}{3!(20-3)!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \]There are 1140 ways to choose 3 packets from 20.

Step 2: Probability of 2 plain packets

To find \( P(2 \text{ plain}) \), we calculate the number of ways to choose 2 plain packets from 10, and 1 non-plain packet from the remaining 10 (maple or apple-cinnamon).Calculate \( \binom{10}{2} \) for choosing 2 plain packets:\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]Calculate \( \binom{10}{1} \) for choosing 1 non-plain packet:\[ \binom{10}{1} = 10 \]The total ways to achieve this combination:\[ 45 \times 10 = 450 \]Thus, the probability is:\[ P(2 \text{ plain}) = \frac{450}{1140} = \frac{15}{38} \approx 0.395 \]

Step 3: Probability of 1 maple and 1 apple-cinnamon

To find \( P(1 \text{ maple, } 1 \text{ apple-cinnamon}) \), we choose 1 maple packet from 6 and 1 apple-cinnamon packet from 4. The third packet can be any of the remaining 18 packets.Calculate \( \binom{6}{1} \) for choosing 1 maple packet:\[ \binom{6}{1} = 6 \]Calculate \( \binom{4}{1} \) for choosing 1 apple-cinnamon packet:\[ \binom{4}{1} = 4 \]Calculate \( \binom{18}{1} \) for choosing the third packet:\[ \binom{18}{1} = 18 \]The total ways to achieve this combination:\[ 6 \times 4 \times 18 = 432 \]Thus, the probability is:\[ P(1 \text{ maple, } 1 \text{ apple-cinnamon}) = \frac{432}{1140} = \frac{18}{95} \approx 0.189 \]

Step 4: Probability of 2 plain and 1 maple

To find \( P(2 \text{ plain, } 1 \text{ maple}) \), we choose 2 plain packets from 10 and 1 maple packet from 6.Calculate \( \binom{10}{2} \) for choosing 2 plain packets:\[ \binom{10}{2} = 45 \]Calculate \( \binom{6}{1} \) for choosing 1 maple packet:\[ \binom{6}{1} = 6 \]The total ways to achieve this combination:\[ 45 \times 6 = 270 \]Thus, the probability is:\[ P(2 \text{ plain, } 1 \text{ maple}) = \frac{270}{1140} = \frac{9}{38} \approx 0.237 \]

Step 5: Probability of 1 of each flavor

To find \( P(1 \text{ of each flavor}) \), we need to choose 1 plain packet from 10, 1 maple packet from 6, and 1 apple-cinnamon packet from 4.Calculate \( \binom{10}{1} \) for choosing 1 plain packet:\[ \binom{10}{1} = 10 \]Calculate \( \binom{6}{1} \) for choosing 1 maple packet:\[ \binom{6}{1} = 6 \]Calculate \( \binom{4}{1} \) for choosing 1 apple-cinnamon packet:\[ \binom{4}{1} = 4 \]The total ways to achieve this combination:\[ 10 \times 6 \times 4 = 240 \]Thus, the probability is:\[ P(1 \text{ of each flavor}) = \frac{240}{1140} = \frac{4}{19} \approx 0.211 \]

Key Concepts

Combination Formula

The combination formula is a fundamental concept in probability and combinatorics, often used when the order of selection does not matter. This is key when calculating probabilities, as it helps determine how many ways there are to pick a subset from a larger set. The formula is expressed in mathematical terms as: \( \binom{n}{r} \), or \( C(n, r) \), and it calculates the number of combinations of \( n \) items taken \( r \) at a time:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Here:

• \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \).
• \( r \) is the number of items to choose, and \( n-r \) is the number of items not chosen.

This formula is powerful because it allows us to compute combinations without listing all possibilities, which is particularly useful when the numbers become large. In our oatmeal packet example, the combination formula helps us determine how many different sets of 3 packets can be chosen from the total 20 available packets without considering the sequence of selection.

Discrete Mathematics

Discrete mathematics is a branch of mathematics that deals with countable, distinct elements. It forms the mathematical foundation for concepts such as probability theory, combinatorics, and computer algorithms. In the context of probability and combinations, discrete mathematics helps us understand discrete events, such as the choice of specific oatmeal packets. Since each packet has a distinct flavor and can either be selected or not, the problem is inherently discrete. Discrete mathematics provides tools to solve problems systematically:

• Counting techniques such as permutations and combinations, which are crucial for understanding possible outcomes.
• Graph theory, number theory, and set theory, which contribute to a broader understanding of structures and relationships in various mathematical problems.
• Logical reasoning, which is indispensable for forming and proving mathematical arguments.

By using discrete mathematics, we gain insights into the ways we can model real-world problems using mathematical objects that possess distinct, separable states, akin to the task of distinguishing between different types of oatmeal packets in a probability problem.

Binomial Coefficients

Binomial coefficients, often associated with the combination formula, form an essential aspect of probability and algebra. They are the coefficients that occur in the expansion of a binomial expression raised to a power, specifically in the expression \((a+b)^n\). The binomial coefficient \(\binom{n}{r}\) corresponds to the term in the expansion where \( a \) is raised to the power of \( n-r \) and \( b \) to the power of \( r \). Binomial coefficients have properties closely tied with Pascal’s Triangle, where each number is the sum of the two directly above it. Key properties of binomial coefficients include:

• Symmetry: \( \binom{n}{r} = \binom{n}{n-r} \)
• Summation identity: \( \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \)

In probability, these coefficients tell us how many ways we can choose \( r \) successes in \( n \) trials, aligning perfectly with our oatmeal example, where we select certain flavors from a broader selection. Binomial coefficients bridge the gap between algebra and combinatorial enumeration, making them invaluable tools in discrete probability scenarios.