A box contains 4 red and 3 blue identical balls. If two are picked at random, one after the other without replacement, find the probability that one is red and the other is blue.
\(\frac{4}{7}\)
\(\frac{2}{7}\)
\(\frac{1}{7}\)
\(\frac{1}{12}\)
Correct answer is A
P(one blue, other red) = P(1st red then blue) or P(1st blue then red)
= \((\frac{4}{7} \times \frac{3}{6}) + (\frac{3}{7} \times \frac{4}{6})\)
= \(\frac{2}{7} + \frac{2}{7} = \frac{4}{7}\)