The number of favorable outcomes (2 red and 2 blue) is: - Dyverse

Understanding the Context

In many probability problems, colorful outcomes are used to represent distinct categories or events. For example:

• A six-sided spinner with two red segments and two blue segments, plus two green (for a total of six),
  - A deck of cards simplified for study with even color representation,
  - Or a sampling experiment involving categorical data.

When we say the number of favorable outcomes is tied to 2 red and 2 blue, we usually mean a scenario where four specific outcomes count as “favorable,” such as landing on certain colors, matching specific pattern arrangements, or satisfying defined conditions.

Calculating Favorable Outcomes for 2 Red and 2 Blue

Key Insights

Instead of a fixed number, the count depends on the total experimental set. Let’s walk through common examples:

Example 1: Random Spin with Color-Based Tiles

Suppose you spin a standard spinner with 6 segments:
- 2 red
- 2 blue
- 2 yellow

If “favorable” means landing on red or blue (i.e., 2 red + 2 blue = 4 segments), then favorable outcomes = 4.

But if “favorable” strictly means only red and blue (4 segments), and yellow (2) are non-favorable, the number is clearly 4.

Final Thoughts

Example 2: Independent Trials – Drawing with Replacement

Imagine repeatedly spinning such a spinner. If you perform 4 independent spins and define a "favorable outcome" as getting red or blue each time, the probability depends on the chance of red or blue per spin:

• Probability of red = ( \frac{2}{6} = \frac{1}{3} )
  - Probability of blue = ( \frac{2}{6} = \frac{1}{3} )
  - Probability of favorable outcome (red or blue) = ( \frac{1}{3} + \frac{1}{3} = \frac{2}{3} )
  - Over 4 trials, expected favorable outcomes = ( 4 \ imes \frac{2}{3} = \frac{8}{3} \approx 2.67 )

But here, favorable outcomes are intended to count as individual events during trials—not cumulative across trials.

Example 3: Counting Permutations

If you randomly sample 4 colored chips from a bag containing 2 red, 2 blue, and 2 green, how many ways can you pick exactly 2 red and 2 blue?