Problem 53 - Algebra and Trigonometry

Question

An urn contains 7 white balls and 3 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 10 balls: (a) If 2 balls are white and 1 is red? (b) If all 3 balls are white? (c) If all 3 balls are red?

Step-by-Step Solution

Answer

(a) 63, (b) 35, (c) 1

Step 1: Understanding the Problem

An urn contains 7 white balls and 3 red balls. Three balls are selected. Calculate the number of ways to draw the balls in each scenario: (a) 2 white and 1 red, (b) all white, (c) all red.

Step 1 - Calculate Total Combinations for Part (a)

Using the combination formula \ C(n, k) = \frac{n!}{k!(n - k)!} \, calculate the number of ways to choose 2 white balls out of 7 and 1 red ball out of 3. Find each individually and multiply the results. \[ C(7, 2) = \frac{7!}{2!(7 - 2)!} = \frac{7 \times 6}{2 \times 1} = 21 \ C(3, 1) = \frac{3!}{1!(3 - 1)!} = \frac{3}{1} = 3 \ \text{Total ways} = 21 \times 3 = 63 \]

Step 2 - Calculate Total Combinations for Part (b)

Using the combination formula, calculate the number of ways to choose 3 white balls out of 7. \[ C(7, 3) = \frac{7!}{3!(7 - 3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

Step 3 - Calculate Total Combinations for Part (c)

Using the combination formula, calculate the number of ways to choose 3 red balls out of 3. \[ C(3, 3) = \frac{3!}{3!(3 - 3)!} = \frac{3 \times 2 \times 1}{3 \times 2 \times 1 \times 1} = 1 \]

Key Concepts

combination formula

The combination formula is essential in combinatorics. It helps us determine the number of ways to choose items from a larger set. The formula is written as: \[ C(n, k) = \frac{n!}{k!(n - k)!} \] Here, \(n\) represents the total number of items, and \(k\) is the number of items to choose. The exclamation mark (!), known as the factorial, means multiplying a number by all positive integers less than itself.
For example:

• \( n! = n \times (n-1) \times (n-2)... \times 1 \).

Using this formula, we can easily determine how many different ways we can select a subset of items from a larger set, which is key in solving our exercise problem.

probability

Probability is about determining how likely an event is to happen. It ranges from 0 (impossible) to 1 (certain). To calculate the probability, we use:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
In the context of our exercise, if we wanted to calculate the probability of selecting 2 white and 1 red ball, we'd first find the number of ways to achieve this (which we've calculated using the combination formula) and then divide it by the total number of ways to select any 3 balls out of the 10. Thus:

• The probability of drawing 2 white and 1 red ball: \( P(2W,1R) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

selection of balls

When selecting balls from a set, we consider different combinations of white and red balls. Let's break it down:

• (a) **2 white and 1 red**: We calculate the ways to choose 2 white from 7 \( C(7, 2) \) and 1 red from 3 \( C(3, 1) \), then multiply them.
  \[ \text{Ways to choose 2 white} = C(7, 2) = 21 \]
  \[ \text{Ways to choose 1 red} = C(3, 1) = 3 \]
  Thus, total ways = 21 * 3 = 63, as already calculated.
• (b) **All 3 white**: Here, we only choose 3 from 7:
  \[ C(7, 3) = 35. \]
• (c) **All 3 red**: Here, since we only have 3 red balls, we choose all of them:
  \[ C(3, 3) = 1. \]

Understanding these combinations helps in tackling more complex probability problems and builds a solid foundation for combinatorics.