NIMCET 2019 Question Paper with solutions

The man is one step away from starting point after 11 steps. This can happen in two ways
(i) He takes ft steps forward and t> steps backward.
(ii) He takes 6 steps forward and ft steps backwards.
∴ Probability of first case${=}^{11}{\mathit{C}}_2{\left(0.4\right)}^5{\left(0.6\right)}^6$
Probability of second ease ${=}^{11}{\mathit{C}}_2{\left(0.4\right)}^6{\left(0.6\right)}^5$
Hence required probability
${=}^{11}{\mathit{C}}_5{\left(0.4\right)}^5{\left(0.6\right)}^6{+}^{11}{\mathit{C}}_6{\left(0.4\right)}^6{\left(0.6\right)}^5$
${=}^{11}{\mathit{C}}_5{\left(0.4\right)}^5{\left(0.6\right)}^5{\left(0.6+0.4\right)}^5$
${=}^{11}{\mathit{C}}_5{\left(0.24\right)}^5=462{\left(0.24\right)}^5$