<p>A man takes a step forward with probability 0.4 and backward with probability 0.6. Then find the probability that at the end of eleven steps he is one step away from the starting point.</p>
<p>\(462 \times (0.24)^5\)</p>
<p>\(462 \times (0.4 \times 0.6)^5\)</p>
<p>\(462 \times (0.4)^6 \times (0.6)^5\)</p>
<p>\(462 \times (0.6)^6 \times (0.4)^5\)</p>
Step-by-Step Solution
Key Concept: The man ends one step away from start when he takes either 6 forward and 5 backward steps (net +1) or 5 forward and 6 backward steps (net -1). Use binomial probability to find each case and sum them.
<p><strong>Step 1:</strong> Identify outcomes. After 11 steps, to be one step away from start, the man must be at position +1 or -1.</p><p><strong>Step 2:</strong> For position +1: He needs 6 forward steps and 5 backward steps. Number of ways = C(11,6). Probability = C(11,6) × (0.4)⁶ × (0.6)⁵</p><p><strong>Step 3:</strong> For position -1: He needs 5 forward steps and 6 backward steps. Number of ways = C(11,5). Probability = C(11,5) × (0.4)⁵ × (0.6)⁶</p><p><strong>Step 4:</strong> Note that C(11,6) = C(11,5), so total probability = 2 × C(11,5) × (0.4)⁵ × (0.6)⁶</p><p><strong>Step 5:</strong> C(11,5) = 462. Calculate: 2 × 462 × (0.4)⁵ × (0.6)⁶ = 924 × (0.01024) × (0.046656) = 924 × 0.0004775... ≈ 0.4406 or 462 × (0.4)⁵ × (0.6)⁶ × 2</p><p><strong>Step 6:</strong> Simplifying: = 2 × C(11,5) × (2/5)⁵ × (3/5)⁶ = 2 × 462 × (32/3125) × (729/15625) = <strong>C(11,5) × 2⁶ × 3⁶/5¹¹</strong></p><p>∴ Answer: A</p>
Correct Answer: A