An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is

We are told that the experiment “succeeds twice as often as it fails.” Let the probability of one success be $$p$$ and the probability of one failure be $$q$$. The statement “twice as often” translates to the ratio

$$p : q = 2 : 1.$$ So we have the relationship

$$p = 2q.$$

Because every individual trial must result in either success or failure, the two probabilities add up to $$1$$. Hence

$$p + q = 1.$$

Substituting $$p = 2q$$ into this equation gives

$$2q + q = 1 \;\;\Longrightarrow\;\; 3q = 1 \;\;\Longrightarrow\;\; q = \frac{1}{3}.$$

Now we find $$p$$ using $$p = 2q$$:

$$p = 2 \left(\frac{1}{3}\right) = \frac{2}{3}.$$

Thus, for each independent trial,

$$p = \frac{2}{3}, \quad q = \frac{1}{3}.$$

We perform $$n = 6$$ independent trials. Let the random variable $$X$$ denote the number of successes in these six trials. Because each trial is independent and has the same success probability, $$X$$ follows the binomial distribution. The probability mass function of a binomially distributed variable is

$$P(X = r) = \binom{n}{r} p^{\,r} q^{\,n-r},$$

where $$\binom{n}{r}$$ is the binomial coefficient.

The question asks for the probability of “at least 5 successes,” which means either 5 successes or 6 successes. We therefore need

$$P(X \ge 5) = P(X = 5) + P(X = 6).$$

First we compute $$P(X = 5)$$. Using the formula with $$n = 6$$ and $$r = 5$$:

$$P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^{6-5}.$$

The binomial coefficient is

$$\binom{6}{5} = 6.$$ Therefore

$$P(X = 5) = 6 \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \cdot \frac{2^5}{3^5} \cdot \frac{1}{3} = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = 6 \cdot \frac{32}{729} = \frac{192}{729}.$$

Next, we compute $$P(X = 6)$$ with $$r = 6$$:

$$P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^{6-6}.$$

Since $$\binom{6}{6} = 1$$ and $$\left(\frac{1}{3}\right)^0 = 1,$$ we get

$$P(X = 6) = 1 \cdot \left(\frac{2}{3}\right)^6 = \frac{2^6}{3^6} = \frac{64}{729}.$$

Now we add the two probabilities:

$$P(X \ge 5) = \frac{192}{729} + \frac{64}{729} = \frac{256}{729}.$$

Hence, the correct answer is Option D.