Problem Analysis:

• Three bags (Bag I, Bag II, Bag III) contain 10 balls each.
• The distribution of Red, Blue, and Green balls varies across the bags.
• A bag is chosen randomly, meaning each bag has a probability of $1/3$ of being selected.
• A ball is drawn from the chosen bag.
• We need to calculate $p$, the probability that a Red ball came from Bag I, and $q$, the probability that a Green ball came from Bag III.
• The final goal is to compute the value of the expression $\left( \frac{1}{p} + \frac{1}{q} \right)$.

Bag Composition Details

The number of balls of each color in the three bags is:

Calculating Probability $p$

Let $B_1$ denote the event of selecting Bag I, and $R$ denote the event of drawing a Red ball.

We need to find $p = P(B_1 | R)$, the probability the ball is from Bag I given it is Red.

Using Bayes' Theorem:

• Probability of selecting Bag I: $P(B_1) = \frac{1}{3}$.
• Probability of drawing a Red ball given Bag I was selected: $P(R | B_1) = \frac{3}{10}$.
• Total probability of drawing a Red ball, $P(R)$, is calculated as:
$ P(R) = P(R | B_1)P(B_1) + P(R | B_2)P(B_2) + P(R | B_3)P(B_3) $ $ P(R) = \left(\frac{3}{10} \times \frac{1}{3}\right) + \left(\frac{4}{10} \times \frac{1}{3}\right) + \left(\frac{5}{10} \times \frac{1}{3}\right) $ $ P(R) = \frac{1}{3} \left( \frac{3}{10} + \frac{4}{10} + \frac{5}{10} \right) = \frac{1}{3} \times \frac{12}{10} = \frac{4}{10} $
• Now, substitute these values into Bayes' Theorem for $p$:
$ p = \frac{\frac{3}{10} \times \frac{1}{3}}{\frac{4}{10}} = \frac{\frac{1}{10}}{\frac{4}{10}} = \frac{1}{4} $

From $p = \frac{1}{4}$, we get $\frac{1}{p} = 4$.

Calculating Probability $q$

Let $B_3$ denote the event of selecting Bag III, and $G$ denote the event of drawing a Green ball.

We need to find $q = P(B_3 | G)$, the probability the ball is from Bag III given it is Green.

Using Bayes' Theorem:

• Probability of selecting Bag III: $P(B_3) = \frac{1}{3}$.
• Probability of drawing a Green ball given Bag III was selected: $P(G | B_3) = \frac{4}{10}$.
• Total probability of drawing a Green ball, $P(G)$, is calculated as:
$ P(G) = P(G | B_1)P(B_1) + P(G | B_2)P(B_2) + P(G | B_3)P(B_3) $ $ P(G) = \left(\frac{5}{10} \times \frac{1}{3}\right) + \left(\frac{3}{10} \times \frac{1}{3}\right) + \left(\frac{4}{10} \times \frac{1}{3}\right) $ $ P(G) = \frac{1}{3} \left( \frac{5}{10} + \frac{3}{10} + \frac{4}{10} \right) = \frac{1}{3} \times \frac{12}{10} = \frac{4}{10} $
• Now, substitute these values into Bayes' Theorem for $q$:
$ q = \frac{\frac{4}{10} \times \frac{1}{3}}{\frac{4}{10}} = \frac{\frac{4}{30}}{\frac{4}{10}} = \frac{1}{3} $

From $q = \frac{1}{3}$, we get $\frac{1}{q} = 3$.

Final Calculation Result

The problem asks for the value of $\left( \frac{1}{p} + \frac{1}{q} \right)$.

Using the calculated values:

The final value is 7.