We are given that $P(A \cup B) = 0.7$ and $P(A \cup B^c) = 0.9$. We need to find $P(A)$.

1. Problem Description

We are given that $P(A \cup B) = 0.7$ and $P(A \cup B^c) = 0.9$. We need to find $P(A)$.

2. Solution Steps

We know that

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Also,

$P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c)$.

Since $P(B^c) = 1 - P(B)$, we can rewrite the second equation as

$P(A \cup B^c) = P(A) + 1 - P(B) - P(A \cap B^c)$.

We also know that $A = (A \cap B) \cup (A \cap B^c)$, and these are disjoint events.

So, $P(A) = P(A \cap B) + P(A \cap B^c)$, and $P(A \cap B^c) = P(A) - P(A \cap B)$.

Now, we can write the second given equation as

$P(A \cup B^c) = P(A) + 1 - P(B) - (P(A) - P(A \cap B))$.

Simplifying this gives

$P(A \cup B^c) = 1 - P(B) + P(A \cap B)$.

Thus, we have the following equations:

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.7$

$P(A \cup B^c) = 1 - P(B) + P(A \cap B) = 0.9$

Adding these two equations gives:

$P(A \cup B) + P(A \cup B^c) = P(A) + P(B) - P(A \cap B) + 1 - P(B) + P(A \cap B) = 0.7 + 0.9$

$P(A) + 1 = 1.6$

$P(A) = 1.6 - 1 = 0.6$

3. Final Answer

$P(A) = 0.6$