Probability
Classical Probability
Grade 12
Question:
<p>A car is parked among <i>N</i> cars standing in a row, but not at either end. On his return, the owner finds that exactly '<i>r</i>' of the <i>N</i> places are still occupied. The probability that the places neighboring his car are empty is</p>
<p>(1) \(\dfrac{(r-1)!}{(N-1)!}\)</p>
<p>(2) \(\dfrac{(r-1)!(N-r)!}{(N-1)!}\)</p>
<p>(3) \(\dfrac{(N-r)(N-r-1)}{(N+1)(N+2)}\)</p>
<p>(4) \(\dfrac{N-r_{C_2}}{N-1_{C_2}}\)</p>
Step-by-Step Solution
Key Concept: Use conditional probability: given that exactly r places are occupied among N total places, find the probability that both neighbors of the owner's car (which must be occupied since it's one of the r occupied places) are empty. This requires recognizing that we need P(both neighbors empty | owner's car occupied and exactly r total occupied).
<p><strong>Step 1:</strong> Let the owner's car be at position k (where 1 < k < N, since it's not at either end).</p><p><strong>Step 2:</strong> Given: Exactly r of N places are occupied (including the owner's car). So (r-1) other cars occupy the remaining (N-1) positions.</p><p><strong>Step 3:</strong> Total ways to place (r-1) cars in (N-1) positions = C(N-1, r-1)</p><p><strong>Step 4:</strong> For both neighboring places to be empty, the (r-1) cars must be placed in the remaining (N-3) positions (excluding position k and its two neighbors). Ways to do this = C(N-3, r-1)</p><p><strong>Step 5:</strong> Required probability = C(N-3, r-1) / C(N-1, r-1)</p><p><strong>Step 6:</strong> Simplifying: [C(N-3, r-1)] / [C(N-1, r-1)] = [(N-3)!(r-1)!(N-r-2)!] / [(N-r-2)!(r-1)!(N-4)!] × [(N-1-r+1)!(r-1)!(N-1)!] = [(N-r-2)(N-r-1)] / [(N-2)(N-1)]</p><p><strong>Step 7:</strong> This simplifies to: (N-r-1)(N-r) / [(N-1)(N-2)]</p><p>∴ Answer: (N-r)(N-r-1) / [(N-1)(N-2)]</p>
Correct Answer: 4