<p><strong>For Problems 19–21:</strong> A player tosses a coin and scores one point for every head and two points for every tail that turns up. He plays on until his score reaches or passes \(n\). \(P_n\) denotes the probability of getting a score of exactly \(n\).</p><p>The value of \(P_n + (1/2)P_{n-1}\) is equal to</p>
Step-by-Step Solution
Key Concept: Set up a recurrence relation by considering how a score of exactly n can be reached: either from score (n-1) by getting a head (1 point) or from score (n-2) by getting a tail (2 points). This creates a functional equation that relates P_n, P_{n-1}, and P_{n-2}.
<p><strong>Step 1:</strong> Set up the recurrence relation. To reach exactly score n, the player must either:</p><ul><li>Be at score (n-1) and flip a head (probability 1/2)</li><li>Be at score (n-2) and flip a tail (probability 1/2)</li></ul><p>Therefore: P_n = (1/2)P_{n-1} + (1/2)P_{n-2}</p><p><strong>Step 2:</strong> Rearrange the recurrence relation:</p><p>P_n = (1/2)P_{n-1} + (1/2)P_{n-2}</p><p>P_n - (1/2)P_{n-1} = (1/2)P_{n-2}</p><p><strong>Step 3:</strong> Shift the index by replacing n with (n+1):</p><p>P_{n+1} - (1/2)P_n = (1/2)P_{n-1}</p><p><strong>Step 4:</strong> Rearrange to find P_n + (1/2)P_{n-1}:</p><p>From Step 1: (1/2)P_{n-1} = P_n - (1/2)P_{n-2}</p><p>Therefore: P_n + (1/2)P_{n-1} = P_n + P_n - (1/2)P_{n-2} = 2P_n - (1/2)P_{n-2}</p><p>Alternatively, using P_{n+1} = (1/2)P_n + (1/2)P_{n-1}, we get:</p><p>∴ <strong>P_n + (1/2)P_{n-1} = (1/2)P_{n-2}</strong> or equivalently <strong>P_{n+1}</strong></p>
Correct Answer: C