A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag.
Find the probability that one ball is red and one is green.

Solution(By Examveda Team)

Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.
$$\eqalign{ & P\left( A \right) = \left( {\frac{5}{8}} \right) \times \left( {\frac{6}{{10}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{8}\,{\text{and}} \cr & P\left( B \right) = \left( {\frac{3}{8}} \right) \times \left( {\frac{4}{{10}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{20}} \cr} $$
Hence, required probability,
$$\eqalign{ & = P(A) + P(B) \cr & = \frac{3}{8} + \frac{3}{{20}} \cr & = \frac{{21}}{{40}} \cr} $$