Q85 / 180 · KCET 2015 · Mathematics

A man takes a step forward with probability \( 0.4 \) and one step backward with probability \( 0.6 \),
then the probability that at the end of eleven steps he is one step away from the starting point
is

  • A
    \(^{11} C_{5} \times(0.48)^{5} \)
  • Correct answer
    B
    \(^{11} C_{6} \times(0.24)^{5} \)
  • C
    \(^{11} C_{5} \times(0.12)^{5} \)
  • D
    \(^{11} C_{6} \times(0.72)^{6} \)

Solution

Let, $p=$ Probability that the man goes forward $=0.4$ And $q=$ Probability that the man goes backward $=0.6$ So, $p+q=1$ Required event is, $5 p, 6 q$ or $6 p, 5 q$ So, required probability is given by ${ }^{n} C_{5} p^{5} q^{6}+{ }^{11} C_{6} p^{6} q^{5}$ $={ }^{11} C_{6} p^{5} q^{5}(q+p)={ }^{11} C_{6}(p q)^{5}={ }^{11} C_{6}(0.24)^{5}$

Hence, the correct answer is option: 'B'.

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