1. **Stating the problem:** We are given the equation $$P(A \cap B) + P(A) + P(B) \times P(A) = 1$$ and the information that events $A$ and $B$ are independent. 2. **Recall the independence rule:** For independent events, $$P(A \cap B) = P(A) \times P(B)$$. 3. **Substitute the independence condition into the equation:** Replace $P(A \cap B)$ with $P(A)P(B)$: $$P(A)P(B) + P(A) + P(B)P(A) = 1$$ 4. **Combine like terms:** Note that $P(A)P(B)$ appears twice: $$P(A)P(B) + P(B)P(A) + P(A) = 1$$ $$2P(A)P(B) + P(A) = 1$$ 5. **Factor out $P(A)$:** $$P(A)(2P(B) + 1) = 1$$ 6. **Solve for $P(A)$:** $$P(A) = \frac{1}{2P(B) + 1}$$ 7. **Interpretation:** The probability $P(A)$ depends on $P(B)$ as above, given the independence and the original equation. **Final answer:** $$\boxed{P(A) = \frac{1}{2P(B) + 1}}$$