A bag contains 2 red and 4 green sweets of the same size and shape. Two boys pick a sweet each from the box, one after the other, without replacement. What is the probability that at least a sweet with green wrapper is picked?
\(\frac{1}{5}\)
\(\frac{2}{5}\)
\(\frac{8}{15}\)
\(\frac{14}{15}\)
Correct answer is D
P(at least one green wrapper) = 1 - P(no green wrapper)
= \(1 - (\frac{2}{6} \times \frac{1}{5})\)
= \(1 - \frac{1}{15}\)
= \(\frac{14}{15}\)