Experiment succeeds twice as often as it fails.
$\therefore \quad$ According to the given condition, if ' $\mathrm{p}$ ' is success and ' $q$ ' is failure, then $p=2 q$
$$
\begin{aligned}
\therefore \quad \mathrm{p}+\mathrm{q}=1 & \Rightarrow 2 \mathrm{q}+\mathrm{q}=1 \\
& \Rightarrow \mathrm{q}=\frac{1}{3} \text { and } \mathrm{p}=\frac{2}{3}
\end{aligned}
$$
Here, $\mathrm{n}=6$
Let $\mathrm{X}$ be the random variable
$$
\therefore \quad \mathrm{X} \sim \mathrm{B}(\mathrm{n}, \mathrm{p})
$$
$\therefore \quad$ Required probability
$$
\begin{aligned}
& =\mathrm{P}(\mathrm{X} \geq 4) \\
& =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \\
& ={ }^6 \mathrm{C}_4 \mathrm{p}^4 \mathrm{q}^2+{ }^6 \mathrm{C}_5 \mathrm{p}^5 \mathrm{q}+{ }^6 \mathrm{C}_6 \mathrm{p}^6 \\
& =15 \times\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^2+6\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)^6 \\
& =\frac{496}{729}
\end{aligned}
$$