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.

Answer: Option D

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.
$$\begin{array}{rl}& P\left(A\right)=\left(\frac{5}{8}\right)\times \left(\frac{6}{10}\right)\\ & =\frac{3}{8}\text{and}\\ & P\left(B\right)=\left(\frac{3}{8}\right)\times \left(\frac{4}{10}\right)\\ & =\frac{3}{20}\end{array}$$
Hence, required probability,
$$\begin{array}{rl}& =P(A)+P(B)\\ & =\frac{3}{8}+\frac{3}{20}\\ & =\frac{21}{40}\end{array}$$