Question

A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps he is one step away from the starting point is

• A. \(\dfrac{2^5\cdot3^5}{5^{10}}\)
• B. \(462\times\left(\dfrac{6}{25}\right)^5\)
• C. \(231\times\dfrac{3^5}{5^{10}}\)
• D. None of these

Hint:

Net displacement problems use binomial distribution.

Answer 1

The Correct Option is B