NIMCET 2013 Question Paper with solutions

Suppose probability of success is 2p and failure is p.
Then, $2\mathit{p}+\mathit{p}=1\Rightarrow \mathit{p}=\frac{1}{3}$
Hence, probability of getting success is $\frac{2}{3}$ and failure is $\frac{1}{3}$ .
∴ Required probability = P (atleast 4 success)
${=}^6{\mathit{C}}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2{+}^6{\mathit{C}}_5{\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^1{+}^6{\mathit{C}}_6{\left(\frac{2}{3}\right)}^6$
$=\frac{15\times 16+6\times 32+64}{3^6}$ $=\frac{16\left(15+12+4\right)}{3^6}$ $=\frac{496}{729}$