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KCET · Maths · Three Dimensional Geometry

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

  1. A \(^{11} C_{5} \times(0.48)^{5} \)
  2. B \(^{11} C_{6} \times(0.24)^{5} \)
  3. C \(^{11} C_{5} \times(0.12)^{5} \)
  4. D \(^{11} C_{6} \times(0.72)^{6} \)
Verified Solution

Answer & Solution

Correct Answer

(B) \(^{11} C_{6} \times(0.24)^{5} \)

Step-by-step Solution

Detailed explanation

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
\({ }^{n} C_{5} p^{5} q^{6}+{ }^{11} C_{6} p^{6} q^{5}\)
\(={ }^{11} C_{6} p^{5} q^{5}(q+p)={ }^{11} C_{6}(p q)^{5}={ }^{11} C_{6}(0.24)^{5}\)