Total outcomes when a fair coin is tossed 3 times:

\[ n(S) = 2^3 = 8 \]

Event \(A\): exactly 2 heads = {HHT, HTH, THH}

\[ |A| = 3 \]

Event \(B\): at most 2 tails = all outcomes except {TTT}

\[ |B| = 7 \]

Since every outcome of \(A\) (two heads ⇒ one tail) is included in \(B\), we have:

\[ A \subseteq B \;\;\Rightarrow\;\; A \cup B = B \]

Therefore:

\[ P(A \cup B) = P(B) = \frac{|B|}{8} = \frac{7}{8} \]

Final Answer:

\[ \boxed{\tfrac{7}{8}} \]

Step 1: Define events.
Let: • A = ball drawn from Bag A • B = ball drawn from Bag B • R = ball drawn is Red

Since a bag is chosen at random: $$P(A) = P(B) = \tfrac{1}{2}$$

Step 2: Probability of Red from each bag.
• From Bag A: $$P(R|A) = \tfrac{3}{3+4} = \tfrac{3}{7}$$ • From Bag B: $$P(R|B) = \tfrac{5}{5+6} = \tfrac{5}{11}$$

Step 3: Total probability of Red.
$$P(R) = P(A)P(R|A) + P(B)P(R|B)$$ $$= \tfrac{1}{2}\cdot\tfrac{3}{7} + \tfrac{1}{2}\cdot\tfrac{5}{11}$$ $$= \tfrac{3}{14} + \tfrac{5}{22}$$ $$= \tfrac{33}{154} + \tfrac{35}{154} = \tfrac{68}{154} = \tfrac{34}{77}$$

Step 4: Apply Bayes’ Theorem.
$$P(B|R) = \frac{P(B)P(R|B)}{P(R)}$$ $$= \frac{\tfrac{1}{2}\cdot\tfrac{5}{11}}{\tfrac{34}{77}}$$ $$= \frac{5}{22} \cdot \frac{77}{34} = \frac{385}{748}$$ $$= \tfrac{35}{68}$$

Correct Probability: $\tfrac{35}{68}$

Answer: Option 1