First bag contains 3 red and 5 black balls and second bag contains 6 red and 4 black balls. A ball is drawn…
First bag contains 3 red and 5 black balls and second bag contains 6 red and 4 black balls. A ball is drawn from each bag. The probability that one ball is red and the other is black, is
$\frac{41}{80}$
$\frac{21}{40}$
$\frac{3}{20}$
$\frac{3}{8}$
Solution
Probability of drawing red ball from first bag $=\frac{3}{8}$ and black ball $=\frac{5}{8}$
Similarly probability of drawing red ball from second bag $=\frac{6}{10}$ and black ball $\frac{4}{10}$
$\therefore$ Required probability $=\left(\frac{3}{8} \times \frac{4}{10}\right)+\left(\frac{5}{8} \times \frac{6}{10}\right)=\frac{12+30}{80}=\frac{21}{40}$