Problem 29 - Calculus for Biology and Medicine

Question

A box contains five red and four blue balls. You choose two balls. (a) How many possible selections contain exactly two red balls, how many exactly two blue balls, and how many exactly one of each color? (b) Show that the sum of the number of choices for the three cases in (a) is equal to the number of ways that you can select two balls out of the nine balls in the box.

Step-by-Step Solution

Answer

(a) Exactly two red: 10. Exactly two blue: 6. Exactly one of each: 20. (b) Sum: 36, matching total combinations of choosing two balls from nine.

Step 1: Calculating two red balls

To find the number of ways to choose exactly two red balls, we use the binomial coefficient formula. We can choose 2 red balls out of 5 available in \( \binom{5}{2} \) ways. Calculate: \( \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \).

Step 2: Calculating two blue balls

To find the number of ways to choose exactly two blue balls, we again use the binomial coefficient. We choose 2 blue balls out of 4 available in \( \binom{4}{2} \) ways. Calculate: \( \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \).

Step 3: Calculating one red and one blue ball

To count the ways to choose one red and one blue ball, we need to multiply the number of ways to pick one red ball and one blue ball. Choose 1 red ball from 5 in \( \binom{5}{1} \) ways and 1 blue ball from 4 in \( \binom{4}{1} \) ways. Calculate: \( \binom{5}{1} = 5 \) and \( \binom{4}{1} = 4 \). The total ways are: \( 5 \times 4 = 20 \).

Step 4: Verify total selections

The sum of the number of choices from the steps above is the total number of ways to pick two balls. Add the results from the previous steps: \( 10 + 6 + 20 = 36 \).

Step 5: Verify total combinations

The total number of ways to pick any 2 balls from 9 is calculated using \( \binom{9}{2} \). Calculate: \( \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \).

Step 6: Conclusion

The total number of ways to select two balls (\( 36 \)) from steps 4 and 5 confirms our calculations. The sum of individual cases equals the total combinations, verifying the consistency.

Key Concepts

Binomial Coefficient

Understanding the binomial coefficient is crucial when working with problems involving combinations, like selecting balls from a box. The binomial coefficient, represented as \( \binom{n}{k} \), tells us the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection. This mathematical concept is foundational in combinatorics and has a simple formula: \\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \( n! \) (n factorial) is the product of all positive integers up to \( n \). In the solution, we calculated how many ways to select different colored balls by plugging values into this formula. For example, when selecting two red balls from five, we used \( \binom{5}{2} \), which simplifies to \( \frac{5 \times 4}{2 \times 1} = 10 \). The same logic applies to choosing two blue balls and combinations of red and blue balls.

Probability

Probability in the context of selecting balls from a box refers to the likelihood of choosing a particular combination of items. It's all about understanding how likely you are to get a pair of red balls, two blue balls, or one of each. Each possible event in our exercise—like choosing two red balls or one red and one blue—has a specific number of favorable outcomes, which we calculated using binomial coefficients. To find the probability of each event, you divide the number of favorable outcomes by the total number of possible selections. For instance, if there are 10 ways to choose two red balls and 36 possible selections overall, the probability of selecting two red balls is \( \frac{10}{36} \). By understanding the number of favorable outcomes and the total number of possibilities, you gain insights into the likelihood of drawing specific combinations.

Discrete Mathematics

Discrete mathematics deals with structures that are fundamentally discrete rather than continuous. It's the math of countable, distinct elements and is the backbone of combinatorics. In the context of the ball selection problem, we use discrete mathematics when counting the combinations of balls. Unlike calculus, which deals with continuous change, discrete mathematics is more about finite sets and distinct values.

• Counting principles: We apply these to determine how many ways there are to arrange or combine sets of distinct items.
• Combinatorics: Utilizes counting principles, such as the binomial coefficient, to find the number of ways to select certain items from a larger group.

Through discrete mathematics, we verstehen how to systematically count the potential arrangements or selections in a self-contained problem, demonstrating its usefulness in analyzing finite data or events.
By mastering these concepts, students unlock the power of discrete mathematics to solve problems involving finite elements and gain a structured approach to dealing with problems involving parts that can't be divided into smaller pieces.