Question

A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{Bag 1=}\left(\text{3R+5B}\right)\text{balls} \\ \text{Bag 2=}\left(\text{6R+4B}\right)\mathrm{balls} \\ P\left(\text{one is red and one is black}\right)=P\left(\text{red from bag 1 and black from bag 2}\right)+P\left(\text{red from bag 2 and black from bag 1}\right) \\ =\frac{3}{8}\times \frac{4}{10}+\frac{5}{8}\times \frac{6}{10} \\ =\frac{12}{80}+\frac{30}{80} \\ =\frac{42}{80} \\ =\frac{21}{40} \end{gathered}$$