We want to compute the probability that in a random selection of 4 birds, there is **at least one bird of each color** (red, green, and blue).

["Title:\nCompute the Probability of Selecting at Least One Bird of Each Color (Red, Green, and Blue) — A Step-by-Step Guide", "---", "Meta Description:\nEver wondered what the probability is of picking at least one bird of each color when selecting 4 birds at random? This article walks you through computing that probability using combinatorics, probability theory, and clear examples.", "---", "## Introduction", "Imagine randomly selecting 4 birds from a collection containing red, green, and blue feathers. How likely is it that your selection includes at least one bird of each color? This classic probability question combines combinatorics with probability concepts and is commonly used in statistics, game theory, and real-world applications like quality control and sampling.", "In this article, we break down the problem step-by-step, explaining how to compute the probability of drawing at least one bird of each color (red, green, and blue) when selecting 4 birds at random. We’ll use fundamental counting and probability principles suitable for educators, students, and data enthusiasts.", "---", "## Problem Setup", "- Colors available: Red, Green, Blue\n- Selection size: 4 birds, drawn randomly\n- Goal: Compute the probability that at least one bird of each color is selected.", "Important assumptions:\n- Birds of each color are abundant enough (or sampling with replacement is assumed) so probabilities don’t diminish with each draw.\n- Each bird has an equal chance of being selected.", "---", "## Step 1: Understand the Total Outcomes", "We begin by computing the total number of ways to choose 4 birds given at least one of each color is required.", "Since we must include at least one bird of each color (red, green, blue) and pick 4 birds total, the possible color distributions that satisfy this are limited.", "### Possible color distributions (with counts):", "Let the counts of red, green, and blue birds in the sample be ( r, g, b ) respectively, such that:\n[\nr + g + b = 4 \quad \ ext{and} \quad r \geq 1, , g \geq 1, , b \geq 1\n]", "This restricts us to integer partitions of 4 into 3 positive parts.", "The valid distributions are the permutations of ( (2,1,1) ) — one color appears twice, the others appear once.", "Why? Because if each color appears at least once, minimum counts are 1 each → 3 birds used, leaving 1 extra. That extra must go to one color → one color appears twice, others once. No 3,0,1 or 4,0,0 combinations work since they miss a color.", "So the only valid partition is permutations of ( (2,1,1) ).", "---", "## Step 2: Count Favorable Outcomes", "We now compute how many favorable outcomes satisfy “at least one of each color” using combinatorics.", "### Step 2.1: Choose which color appears twice\nThere are ( \binom{3}{1} = 3 ) ways to choose which color occurs twice (red, green, or blue).", "### Step 2.2: Count ways to assign birds in the sample\nSuppose red appears twice, green once, blue once. The number of distinct arrangements (order of selection) of the multiset {Red, Red, Green, Blue} is:\n[\n\frac{4!}{2!1!1!} = 12\n]\nSo, for each choice of the duplicated color, there are 12 distinct orderings.", "Since there are 3 choices for which color is duplicated, total favorable ordered sequences:\n[\n3 \ imes 12 = 36\n]", "Alternative (non-ordering) count:\nWe can also compute using combinations with repetition, but since sampling is assumed without replacement by type in this combinatorial model, the multinomial approach above is appropriate.", "Thus, total favorable outcomes = 36", "---", "## Step 3: Count Total Possible Outcomes", "Now, we calculate total possible ways to select 4 birds, allowing any color distribution (with possible absence), assuming birds of each color are abundant (or sampling is independent and random).", "But here’s a key point: if we treat each bird as distinguishable and sampling with unlimited repetition (or large population), we use the multiset counting approach. However, for probability, a better model is selecting one bird uniformly at random from a well-stocked pool, with equal chance per color at each draw.", "But since exactly “at least one of each of three colors” in 4 draws is rare, we use combinatorics over equally likely color combinations.", "A standard and rigorous model is:", "Assume each bird selected belongs to one of three colors (red, green, blue), with equal probability ( \frac{1}{3} ) per color, independent of others (i.e., sampling with replacement).", "Then total number of possible color sequences for 4 birds:\n[\n3^4 = 81\n]", "This assumes independence and equal likelihood — a common assumption unless population size is specified.", "So total possible outcomes = 81", "---", "## Step 4: Compute Favorable Outcomes (At Least One of Each Color)", "We now count how many of these 81 sequences contain at least one red, one green, and one blue bird.", "Use inclusion-exclusion principle for clean counting.", "Let:\n- ( A ) = set of sequences with no red\n- ( B ) = no green\n- ( C ) = no blue", "We want sequences not in A, B, or C → i.e., with all three colors present.", "Total sequences: ( 3^4 = 81 )", "[\n|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|\n]", "Compute each:", "- ( |A| ): no red → only green and blue → ( 2^4 = 16 )\n- ( |B| ): no green → red and blue → 16\n- ( |C| ): no blue → red and green → 16", "- ( |A \cap B| ): no red, no green → only blue → ( 1^4 = 1 )\n- ( |A \cap C| ): no red, no blue → only green → 1\n- ( |B \cap C| ): no green, no blue → only red → 1", "- ( |A \cap B \cap C| ): no bird of any color → impossible → 0", "So:\n[\n|A \cup B \cup C| = 16 + 16 + 16 - 1 - 1 - 1 + 0 = 48 - 3 = 45\n]", "Therefore, sequences with all three colors:\n[\n81 - 45 = 36\n]", "Matches our earlier enumeration — expected!", "---", "## Step 5: Compute the Probability", "[\nP(\ ext{at least one of each color}) = \frac{\ ext{favorable outcomes}}{\ ext{total outcomes}} = \frac{36}{81} = \frac{4}{9}\n]", "---", "## Final Answer", "[\n\boxed{\frac{4}{9}}\n]", "So, when randomly selecting 4 birds with equal probability of each color (red, green, blue), the probability that at least one bird of each color appears is exactly ( \frac{4}{9} ).", "---", "## Why This Matters", "Understanding such probabilities helps in fields like:", "- Ecology: Estimating biodiversity presence from field samples\n- Quality Control: Detecting defective items from mixed batches\n- Biology: Genetic modeling in population studies\n- Statistics Education: Introducing combinatorics in probability", "---", "## Additional Tips", "- Always clarify whether sampling is with replacement/color-only or one by one from an ensemble.\n- Use symmetry: when multiple categories are equally likely, grouping cases by permutation reduces complexity.\n- The inclusion-exclusion method is powerful for “at least one” problems with overlapping constraints.", "---", "Keywords: probability, at least one bird of each color, compute probability, combinatorics, sampling without replacement, multinomial probability, probability probability, statistical probability, bird color probability, 4 birds, red green blue color, probability calculation, inclusive counting, inclusion-exclusion principle", "---", "Note: For sampling without replacement from a finite population, the method changes — you’d use combinations: number of 4-bird groups containing all three colors divided by total groups. But with large populations or sampling with replacement, the 81 total is often valid.", "---", "Check back here anytime you want to compute probabilities for colorful combinatorics problems!"]