$$ \text { Given three indentical bags each containing } 10 \text { balls, whose colours are as follows : } $$

$$ \begin{array}{lccc} & \text { Red } & \text { Blue } & \text { Green } \\ \text { Bag I } & 3 & 2 & 5 \\ \text { Bag II } & 4 & 3 & 3 \\ \text { Bag III } & 5 & 1 & 4 \end{array} $$

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is $q$, then the value of $\left(\frac{1}{p}+\frac{1}{q}\right)$ is:

<p><strong>Probability that a Red ball comes from Bag I ($p$)</strong>:</p> <p>$ p(B_1 / R) = \frac{p(B_1) \cdot p(R / B_1)}{p(R)} $</p> <p>Substituting the values,</p> <p>$ p(B_1 / R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} $</p> <p>Simplify to find $ p $:</p> <p>$ = \frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4} $</p> <p>So, $ p = \frac{1}{4} $.</p></p> <p><p><strong>Probability that a Green ball comes from Bag III ($q$)</strong>:</p> <p>$ p(B_3 / G) = \frac{p(B_3) \cdot p(G / B_3)}{p(G)} $</p> <p>Substitute the values,</p> <p>$ p(B_3 / G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}} $</p> <p>Simplify to find $ q $:</p> <p>$ = \frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3} $</p> <p>So, $ q = \frac{1}{3} $.</p></p> <p><p><strong>Calculation of $\left(\frac{1}{p} + \frac{1}{q}\right)$:</strong></p> <p>$ \frac{1}{p} = 4, \quad \frac{1}{q} = 3 $</p> <p>Therefore,</p> <p>$ \left(\frac{1}{p} + \frac{1}{q}\right) = 4 + 3 = \boxed{7} $</p></p>