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

• The man would be a step away from starting point in two cases:
  
  1. He takes 6 steps forward and 5 steps backward
  2. He takes 5 steps forward and 6 steps backward
  
  For 1st case:
  p1 = (11C6)(0.4)^6*(0.6)^5
  
  for 2nd case:
  
  p2 = 11C5(0.4)^5*(0.6)^6
  
  total probability = p1 + p2 = 11C6 * (0.4)^6 * (0.6)^5 + 11C5 * (0.4)^5 * (0.6)^6
  = 11C5 * (0.4)^5 * (0.6)^5 * (0.4 + 0.6) >>>>>>>(because 11C6 = 11C5)
  = (11x10x9x8x7)/(1x2x3x4x5) * (0.24)^5
  = 0.368.
• 11 years agoHelpfull: Yes(3) No(2)
• Since after 11 steps he is one away from the starting point, there are 2 ways this can happen:
  a) He will take 5 steps forward and 6 steps backward
  b) He will take 6 steps forward and 5 steps backward
  There can be permutation in these two choices.
  
  The probability for the 1st point: 0.4^5+0.6^6=0.01024+0.046656=0.057(approx)
  The probability for the 2nd point: 0.4^6+0.6^5=0.00409+0.0777=0.08185 (approx)
  SO required probability is: 0.08185+0.057= 0.13885
• 11 years agoHelpfull: Yes(0) No(1)
• ans:-0.4
  initially we are at '0' then start step forward and backward respectively upto 11 steps.now we reach -0.4 position .i.e,this 1 step away to reach starting point.
• 11 years agoHelpfull: Yes(0) No(0)
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