To solve this problem, we need to find $P(B|A)$, which is the conditional probability of $B$ given $A$.

Given:

• $P(A) = \frac{3}{5}$
• $P(A \cup B) = \frac{2}{3}$
• $A$ and $B$ are independent events.

Steps to Solve:

1. Use the formula for the union of two events:

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

   Since $A$ and $B$ are independent:

   $P(A \cap B) = P(A) \cdot P(B)$

   Substituting this into the union formula gives:

   $P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$

2. Substitute the known values:

   $\frac{2}{3} = \frac{3}{5} + P(B) - \left(\frac{3}{5} \cdot P(B)\right)$

   Simplify the equation:

   $\frac{2}{3} = \frac{3}{5} + P(B) \left(1 - \frac{3}{5}\right)$

   $\frac{2}{3} = \frac{3}{5} + P(B) \cdot \frac{2}{5}$

   Multiply everything by 15 to eliminate the denominators:

   $10 = 9 + 6P(B)$

   Subtract 9 from both sides:

   $1 = 6P(B)$

   Therefore, $P(B) = \frac{1}{6}$.

3. Find $P(B|A)$:

   $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A) \cdot P(B)}{P(A)} = P(B)$

   Substituting the value of $P(B)$:

   $P(B|A) = \frac{1}{6}$

Final Answer:

$P(B|A) = \frac{1}{6}$

Would you like more details, or do you have any further questions?

Here are five related questions for practice:

1. If $P(A) = 0.7$ and $P(B) = 0.4$, and $A$ and $B$ are independent, what is $P(A \cap B)$?
2. Given that $P(A) = 0.5$, $P(B) = 0.3$, and $P(A \cup B) = 0.65$, find $P(A \cap B)$.
3. If $P(A \cup B) = 0.9$ and $P(A \cap B) = 0.2$, what is $P(A)$ if $P(B) = 0.4$?
4. What is the conditional probability $P(A|B)$ if $P(B) = 0.2$ and $P(A \cap B) = 0.1$?
5. How would the independence of events $A$ and $B$ affect the calculation of $P(A \cup B)$?

Tip: When dealing with conditional probabilities, remember that the definition $P(B|A) = \frac{P(A \cap B)}{P(A)}$ can simplify many problems, especially when events are independent.