1. **Problem statement:** Given that $P(A|B) + P(B|A) = 0.8$ and $\frac{1}{P(B)} + \frac{1}{P(A)} = 4$, find $P(B \cap A)$. 2. **Recall the conditional probability formulas:** $$P(A|B) = \frac{P(A \cap B)}{P(B)} \quad \text{and} \quad P(B|A) = \frac{P(A \cap B)}{P(A)}$$ 3. Substitute these into the given equation: $$\frac{P(A \cap B)}{P(B)} + \frac{P(A \cap B)}{P(A)} = 0.8$$ 4. Factor out $P(A \cap B)$: $$P(A \cap B) \left( \frac{1}{P(B)} + \frac{1}{P(A)} \right) = 0.8$$ 5. From the problem, we know: $$\frac{1}{P(B)} + \frac{1}{P(A)} = 4$$ 6. Substitute this into the equation: $$P(A \cap B) \times 4 = 0.8$$ 7. Solve for $P(A \cap B)$: $$P(A \cap B) = \frac{0.8}{4} = 0.2$$ **Final answer:** $P(B \cap A) = 0.2$ which corresponds to option ⓑ.