Question

A man takes either step forward with probability 0.6 and a step backward with probability 0.4. then the probability that he is 1 step away from the starting point at the end of 11 steps is

• A. ${ } ^ { 11 } \mathrm { C } _ { 5 } ( 0.2 ) ^ { 5 }$
• B. ${ } ^ { 11 } \mathrm { C } _ { 6 } ( 0.32 ) ^ { 5 }$
• C. ${ } ^ { 11 } \mathrm { C } _ { 5 } ( 0.24 ) ^ { 5 }$
• D. ${ } ^ { 11 } \mathrm { C } _ { 6 } ( 0.24 ) ^ { 6 }$

Answer 1

Answer: (C)

${ } ^ { 11 } \mathrm { C } _ { 5 } ( 0.24 ) ^ { 5 }$

Answer 2

P = 0.6 q = 0.4

Let X be no. of forward steps,

then X = {0,1,2,------11}

Required probability

= P(X=5) + P(X=6)

=

= ${ } ^ { 11 } \mathrm { C } _ { 5 } ( 0.6 ) ^ { 5 } ( 0.4 ) ^ { 5 } \mathrm { x } 1$

= ${ } ^ { 11 } \mathrm { C } _ { 5 } ( 0.24 ) ^ { 5 }$ .