Given that an experiement succeeds twice as often as it fails we need to find the probability that in the next six trials, there will be at least 4 successes.

Since the experiment succeed twice as often it fails \(p=\large\;\frac{2}{3}\)\(;\;q=1\;-\;p=1\;-\large\;\frac{2}{3}=\frac{1}{3}\)

Let X be the random variable that represents the number of successes in $n=6$ trials.

A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $p(X=r)=C^{n}_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$

$P(X=r)=C^{6}_{r} \large(\frac{2}{3})^{r} \frac{1}{3}^{6-r}$,

P (at least 4 successes) = P (X $\geq$ 4) =\(p(\;X\;=\;4\;)\;+\;p(\;X\;+\;5\;)\;+\;p(\;X\;=6\;)\)

=\(\large\;c^6_4(\frac{2}{3})^4\;(\frac{1}{3})^2\;+\;c^6_5\;(\frac{2}{3})^5\;(\frac{1}{3})+c^6_6\;(\frac{2}{3})^6\;(\frac{1}{3})^0\)

=\(\;15\;\large\frac{2^4}{3^6}\;\)\(+\;6\;\large\frac{2^5}{3^6}\;\)\(+\;\large\frac{2^6}{3^6}\)

=\(\large\;\frac{2^4}{3^6}\;\)\([15\;+\;12\;+\;4]\)

=\(\;31\;\large\frac{2^4}{3^6}\)