Problem 32 - Calculus for Biology and Medicine

Question

A bag contains 45 beans of three different varieties. Each variety is represented 15 times in the bag. You grab 9 beans out of the bag. (a) Count the number of ways that each variety can be represented exactly three times in your sample. (b) Count the number of ways that only one variety appears in your sample.

Step-by-Step Solution

Answer

(a) 94,109,375 ways; (b) 15,015 ways.

Step 1: Understand the Problem

We need to calculate the number of ways to select 9 beans such that each variety is represented 3 times. Then, we need to calculate the number of ways to select 9 beans such that only one variety is represented.

Step 2: Calculate Ways for Equal Representation (Part a)

Use combinations to find out how to select 3 beans from each of the 15 beans per variety. The number of ways to choose 3 beans from a variety of 15 is calculated as \( \binom{15}{3} \) for each variety.Therefore, total ways for all varieties: \( (\binom{15}{3})^3 \).

Step 3: Calculate the Combination for One Variety (Part b)

To calculate the number of ways to choose all 9 beans from just one variety, use combinations to select 9 out of 15 beans. This is given by the combination \( \binom{15}{9} \).Since there are 3 varieties, multiply this result by 3 to account for any variety: \( 3 \times \binom{15}{9} \).

Step 4: Compute Values

Compute \( \binom{15}{3} = 455 \) and then \( (\binom{15}{3})^3 = 455^3 \). Similarly, compute \( \binom{15}{9} = 5005 \) and \( 3 \times \binom{15}{9} = 3 \times 5005 \).

Step 5: Final Calculations

Calculate the final numerical answers for both parts:(a) \( 455^3 = 94,109,375 \) ways for equal representation.(b) \( 3 \times 5005 = 15,015 \) ways for just one variety represented.

Key Concepts

Calculation of Combinations

When tackling questions involving selecting items, combinations play a crucial role. In probability, combinations are used when the order of selection does not matter. For example, if you are picking a team of 3 people from a group of 5, it doesn't matter who you pick first or last, the team remains the same.

Mathematically, combinations are represented by the symbol \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. The formula to calculate combinations is:

• \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

Here, "\(!\)" is the factorial operator, denoting the product of all positive integers less than or equal to a number.

In the bean problem, for part (a) where each variety needs to be represented thrice out of 15, we calculate \( \binom{15}{3} \). This calculates the number of ways to choose 3 beans from a variety of 15. To cover all three varieties, we calculate \( (\binom{15}{3})^3 \), implying independence between the selection from each variety.

Probability Theory Problems

In probability theory, it's essential to approach problems with a clear plan for calculating the likelihood of various outcomes. Let's consider our bean bag example. We wanted to determine the number of ways different scenarios can happen: equal representation of varieties or representation of only one variety.

For part (a), equal representation means selecting beans so each variety appears the same number of times. It's a typical combinatorial problem where we need to count the favorable situations among the total possible outcomes. In this case, there are multiple steps within the calculation involving determining how to choose 3 beans from a set of 15 for each of the three varieties.

For part (b), where only one variety should appear, we calculate this by choosing all 9 beans from just one variety. With 3 different sets to choose from, multiplying \( \binom{15}{9} \) by 3 helps find the number of ways to get beans all from a singular variety. These calculations do not provide direct probabilities, but they are the building blocks for understanding them when combined with the concept of total possible outcomes.

Combinatorial Analysis

Combinatorial analysis is the systematic way of counting various configurations or arrangements of items within certain constraints. This method is essentially about understanding how we can arrange or select objects, and it's extremely helpful in solving complex probability problems.

Take our earlier bean problem. First, for each desired arrangement (whether evenly distributed or single-variety), we delineate the conditions and apply the appropriate combination logic. For instance, in the equal distribution scenario, the analysis simplifies by calculating arrangements for each variety purely independently and then bringing them together. This independent calculation and cumulative product represent a unique count of those specific arrangements.

Analyzing this through combinatorics allows you to see that despite having seemingly large datasets, the numbers break down into easily calculable patterns through step-by-step combination logic. These resulting numbers, such as 94,109,375 and 15,015 in our bean exercise, portray the extensive yet concise power of combinatorial methods in simplifying and answering probability-wide questions.