Problem 54 - Algebra and Trigonometry

Question

An urn contains 15 red balls and 10 white balls. Five balls are selected. In how many ways can the 5 balls be drawn from the total of 25 balls: (a) If all 5 balls are red? (b) If 3 balls are red and 2 are white? (c) If at least 4 are red balls?

Step-by-Step Solution

Answer

(a) \[ \binom{15}{5} \], (b) \[ \binom{15}{3} \times \binom{10}{2} \], (c) \[ \binom{15}{4} \times \binom{10}{1} + \binom{15}{5} \].

Step 1: Total Possible Combinations

First, determine the total number of ways to choose 5 balls from 25. This can be found using the combination formula \(\binom{n}{k}\), where \( n \) is the total number of items and \( k \) is the number of items to choose. \[ \text{Total combinations} = \binom{25}{5} \]

Step 2: Combinations for All Red Balls

Calculate the number of ways to select 5 red balls from the 15 available red balls using the combination formula. \[ \binom{15}{5} \]

Step 3: Combinations for 3 Red and 2 White Balls

Calculate the number of ways to select 3 red balls from the 15 and 2 white balls from the 10 using the combination formula for each, then multiply the results. \[ \binom{15}{3} \times \binom{10}{2} \]

Step 4: Combinations for At Least 4 Red Balls

Break this into two cases: one for 4 red and 1 white, and another for all 5 red. Calculate each case separately and then sum them up. \[ \binom{15}{4} \times \binom{10}{1} + \binom{15}{5} \]

Key Concepts

combinatorial analysis

Combinatorial analysis is a branch of mathematics that deals with counting, arranging, and combining objects. It's essential for solving many probability problems.
For example, if you want to know how many ways you can choose or arrange a set of items, you use principles of combinatorial analysis.
In our exercise, we need to calculate different ways to choose red and white balls from an urn.
We use combinatorial analysis because we're dealing with combinations—where the order of the balls doesn't matter—and we want to consider every possible way to draw a specific number of balls.
This approach makes it easier to solve the problem by breaking it into manageable calculations.

combination formula

The combination formula is a key tool in combinatorial analysis. It helps us determine how many ways we can choose a subset of items from a larger set, without considering the order.
The formula is given by:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!} \] This formula is read as 'n choose k,' where n is the total number of items, and k is the number of items to choose.
In our exercise:

• To find how many ways we can choose 5 balls from 25, we use \[\binom{25}{5}\].
• For selecting all 5 red balls from 15 red balls, we use \[\binom{15}{5}\].
• For selecting 3 red and 2 white balls, we use \[\binom{15}{3}\] and \[\binom{10}{2}\] and then multiply the results, since there are \[\binom{15}{3}\] ways to choose 3 red balls and \[\binom{10}{2}\] ways to choose 2 white balls.

Using this formula makes it easier to handle complex problems involving combinations.

probability problems

Probability problems often require us to determine the likelihood of a certain event happening.
To solve these problems, we frequently use the combination formula to find out the number of favorable outcomes and divide it by the total number of possible outcomes.
In our exercise, we're interested in different scenarios, such as all red balls, a combination of red and white balls, or at least a certain number of red balls.
First, we calculate the number of ways to draw balls in each scenario:

• For all 5 red balls: \[\binom{15}{5}\].
• For 3 red and 2 white balls: \[\binom{15}{3} \times \binom{10}{2}\].
• For at least 4 red balls, we calculate two cases: 4 red and 1 white \[\binom{15}{4} \times \binom{10}{1}\], and all 5 red \[\binom{15}{5}\], adding them together.

By breaking down the problem and calculating each scenario, we can confidently determine the number of ways to draw the balls as required.