<p>Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?</p>
Step-by-Step Solution
Key Concept: Recognize that 'at most 6 right-handed' is the complement of '7 or more right-handed', making it easier to sum only 4 terms instead of 7. This binomial probability uses the complement rule: P(X ≤ 6) = 1 - P(X ≥ 7).
<p><strong>Step 1:</strong> Identify the binomial setting. We have n = 10 trials, probability of success (right-handed) p = 0.9, and we want P(X ≤ 6) where X = number of right-handed people.</p><p><strong>Step 2:</strong> Use the complement rule. P(X ≤ 6) = 1 - P(X ≥ 7) = 1 - P(X = 7) - P(X = 8) - P(X = 9) - P(X = 10).</p><p><strong>Step 3:</strong> Express using binomial probability formula. Each term P(X = r) = ${}^{10}C_r (0.9)^r (0.1)^{10-r}$, so the sum runs from r = 7 to r = 10.</p><p><strong>Step 4:</strong> Write the final answer using summation notation.</p><p>∴ Answer: $1 - \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r}$</p>
Correct Answer: \(1 - \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r}\)