An experiment succeeds twice as often as it fails. The probability that in the next six trials, there will be at least 4 successes, is:

Solution:
​$$p = \frac{2}{3}, \quad q = \frac{1}{3}, \quad n = 6 \\ \ \\P(X \geq 4) = P(4) + P(5) + P(6) \\ \ \\= \binom{6}{4} p^4 q^2 + \binom{6}{5} p^5 q + \binom{6}{6} p^6 \\ \ \\= \left( \frac{2}{3} \right)^4 \left[ \frac{15}{9} + \frac{12}{9} + \frac{4}{9} \right] =\\ \ \\ \left( \frac{2}{3} \right)^4 \cdot \frac{31}{9} \\$$​​

Correct answer is (a)