To solve this problem, we need to determine the probability of getting at least four successes in six trials, given that the probability of success is twice the probability of failure.
Step-by-Step Solution:
- Defining the probabilities:
- Let the probability of failure be \(p\) and the probability of success be \(2p\).
- Since these are the only two outcomes possible, we have: \(p + 2p = 1\).
- This simplifies to \(3p = 1\), so \(p = \frac{1}{3}\) and consequently, the probability of success \(= \frac{2}{3}\).
- Understanding the distribution: The problem describes a binomial distribution with the number of trials \(n = 6\) and probability of success \(p_s = \frac{2}{3}\).
- Calculating probability of at least four successes:
- We need the sum of probabilities for having exactly 4, 5, and 6 successes.
- Using the binomial probability formula: \(P(X = k) = \binom{n}{k} p_s^k (1-p_s)^{n-k}\), where \(n\) is the number of trials, \(k\) is the number of successes, and \(p_s\) is the probability of success.
- For 4 successes: \(P(X = 4) = \binom{6}{4} \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2\)
- For 5 successes: \(P(X = 5) = \binom{6}{5} \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right)^1\)
- For 6 successes: \(P(X = 6) = \binom{6}{6} \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^0\)
- Calculating the individual probabilities:
- \(P(X = 4) = \binom{6}{4} \cdot \left( \frac{2}{3} \right)^4 \cdot \left( \frac{1}{3} \right)^2 = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = \frac{240}{729}\)
- \(P(X = 5) = \binom{6}{5} \cdot \left( \frac{2}{3} \right)^5 \cdot \left( \frac{1}{3} \right)^1 = 6 \cdot \frac{32}{243} = \frac{192}{729}\)
- \(P(X = 6) = \binom{6}{6} \cdot \left( \frac{2}{3} \right)^6 = \frac{64}{729}\)
- Adding the probabilities for at least four successes:
- Total probability \(= P(X = 4) + P(X = 5) + P(X = 6)\)
- \(= \frac{240}{729} + \frac{192}{729} + \frac{64}{729} = \frac{496}{729}\)
Thus, the correct answer is \(\frac{496}{729}\).