A bag contains 21 marbles; there are only red marbles and blue marbles in the bag. There are twice as many red marbles as there are blue marbles. If exactly two marbles were to be selected at random, without replacement, what is the probability that one would be red and one would be blue? Express your answer as a common fraction.
+2666 There are 14 red marbles and 7 blue marbles in the bag.
The probability of getting a red marble is \(2 \over 3\). The probability of getting a blue marble is \(7 \over 20\), because we already took 1 marble.
Thus, the probability is \({2 \over 3} \times {7 \over 20} = \color{brown}\boxed{7 \over 30}\)
