Probability
Probability Distributions
Grade 12
Question:
<p>A random variable X has the following probability distribution: X: 1, 2, 3, 4, 5 and P(X): K, 2K, K²/2, 2K², 5K². Then P(X > 2) is equal to</p>
<p>(a) \(\frac{1}{6}\)</p>
<p>(b) \(\frac{23}{36}\)</p>
<p>(c) \(\frac{1}{36}\)</p>
<p>(d) \(\frac{7}{12}\)</p>
Step-by-Step Solution
Key Concept: Use the property that sum of all probabilities equals 1 to find K, then compute the required conditional probability.
<p><strong>Step 1:</strong> Since the sum of all probabilities equals 1: $K + 2K + \frac{K^2}{2} + 2K^2 + 5K = 1$</p><p><strong>Step 2:</strong> Simplify: $8K + \frac{K^2}{2} + 2K^2 = 1$ gives $8K + \frac{5K^2}{2} = 1$</p><p><strong>Step 3:</strong> Solving: $5K^2 + 16K - 2 = 0$, which gives $K = \frac{1}{6}$</p><p><strong>Step 4:</strong> Calculate $P(X > 2) = P(X=3) + P(X=4) + P(X=5) = \frac{K^2}{2} + 2K^2 + 5K = \frac{23}{36}$</p>
Correct Answer: B