Solution: This is a binomial probability problem with $ n = 6 $, $ p = 0.3 $, and we want $ P(X < 3) = P(X=0) + P(X=1) + P(X=2) $.

["# Solving the Binomial Probability Problem: P(X < 3) for n = 6, p = 0.3", "When analyzing discrete probability scenarios, binomial probability problems are among the most common and widely applicable. In this article, we explore how to solve a specific binomial probability problem: determining ( P(X < 3) ) where the random variable ( X ) follows a binomial distribution with parameters ( n = 6 ) and success probability ( p = 0.3 ).", "## Understanding the Binomial Distribution", "The binomial distribution gives the probability of achieving exactly ( k ) successes in ( n ) independent trials, each with a success probability ( p ). The probability mass function is defined as:", "[\nP(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}\n]", "where:\n- ( \binom{n}{k} ) is the binomial coefficient, representing the number of ways to choose ( k ) successes from ( n ) trials,\n- ( p^k ) is the probability of ( k ) successes,\n- ( (1 - p)^{n-k} ) is the probability of ( n - k ) failures.", "## Problem Statement", "We are tasked with computing:", "[\nP(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\n]", "Given:\n- ( n = 6 )\n- ( p = 0.3 )\n- ( 1 - p = 0.7 )", "---", "## Step-by-Step Solution", "### 1. Compute ( P(X = 0) )", "[\nP(X = 0) = \binom{6}{0} (0.3)^0 (0.7)^6 = 1 \cdot 1 \cdot (0.7)^6 = (0.7)^6\n]", "[\n(0.7)^6 = 0.117649\n]", "### 2. Compute ( P(X = 1) )", "[\nP(X = 1) = \binom{6}{1} (0.3)^1 (0.7)^5 = 6 \cdot 0.3 \cdot (0.7)^5\n]", "First calculate ( (0.7)^5 ):", "[\n(0.7)^5 = 0.16807\n]", "Now:", "[\n6 \cdot 0.3 \cdot 0.16807 = 1.8 \cdot 0.16807 = 0.302526\n]", "### 3. Compute ( P(X = 2) )", "[\nP(X = 2) = \binom{6}{2} (0.3)^2 (0.7)^4 = 15 \cdot (0.09) \cdot (0.7)^4\n]", "Calculate ( (0.7)^4 ):", "[\n(0.7)^4 = 0.2401\n]", "Now:", "[\n15 \cdot 0.09 \cdot 0.2401 = 1.35 \cdot 0.2401 = 0.324135\n]", "---", "### 4. Sum the Probabilities", "[\nP(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.117649 + 0.302526 + 0.324135\n]", "[\nP(X < 3) = 0.74431\n]", "---", "## Final Answer", "[\n\boxed{P(X < 3) \approx 0.7443}\n]", "This result indicates there is approximately a 74.43% chance that the number of successes in 6 independent trials with a 30% success probability will be less than 3.", "---", "## Why This Matters", "Understanding and computing binomial probabilities is essential in statistics, quality control, risk analysis, and decision-making under uncertainty. This problem format teaches how to break down compound probabilities into manageable components and apply the binomial formula step by step.", "For those encountering this type of problem, always:\n- Identify the parameters ( n ), ( p ), and ( 1-p ),\n- Use the correct binomial probability formula,\n- Compute powers and binomial coefficients carefully,\n- Sum the individual probabilities to find the cumulative outcome.", "This method provides a solid foundation for tackling more complex probabilistic models."]