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
- A \(^{11} C_{5} \times(0.48)^{5} \)
- B \(^{11} C_{6} \times(0.24)^{5} \)
- C \(^{11} C_{5} \times(0.12)^{5} \)
- D \(^{11} C_{6} \times(0.72)^{6} \)
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
\({ }^{n} C_{5} p^{5} q^{6}+{ }^{11} C_{6} p^{6} q^{5}\)
\(={ }^{11} C_{6} p^{5} q^{5}(q+p)={ }^{11} C_{6}(p q)^{5}={ }^{11} C_{6}(0.24)^{5}\)
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
\({ }^{n} C_{5} p^{5} q^{6}+{ }^{11} C_{6} p^{6} q^{5}\)
\(={ }^{11} C_{6} p^{5} q^{5}(q+p)={ }^{11} C_{6}(p q)^{5}={ }^{11} C_{6}(0.24)^{5}\)
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