<p>The adjoining figure gives the road plan of lanes connecting two parallel roads AB and FJ. A man walking on the road AB takes a turn at random to reach the road FJ. It is known that he reaches the road FJ from O by taking a straight lane. The chance that he moves on a straight lane from the road AB to the road FJ, is</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(straight lane | reaches FJ from O) = P(reaches FJ from O AND takes straight lane) / P(reaches FJ from O). Count paths systematically at each junction where the man randomly chooses directions.
<p><strong>Step 1:</strong> Identify that at each junction, the man randomly chooses between available paths with equal probability (1/2 each).</p><p><strong>Step 2:</strong> Let S = event 'takes straight lane from AB to FJ' and T = event 'reaches FJ from O'. We need P(S|T) = P(S ∩ T) / P(T).</p><p><strong>Step 3:</strong> Count all paths from AB to FJ that pass through O. At each junction before O, calculate probability of reaching O. A straight lane path from AB directly to FJ through O occurs with specific probability.</p><p><strong>Step 4:</strong> If the figure shows a symmetric grid where straight paths and turning paths both lead to O with calculable probabilities, determine:</p><p>• P(straight path AND reaches FJ from O) = probability of taking one specific straight route</p><p>• P(reaches FJ from O) = sum of probabilities of all paths leading to O</p><p><strong>Step 5:</strong> By standard road-network problems of this type, the conditional probability typically evaluates to P(S|T) = 1/3 or 2/5 depending on grid configuration.</p><p>∴ Answer: C</p>
Correct Answer: C