Given: P(X Y) = (11)/(15) and P(X) = (1)/(3).

Another one Faith — let's solve it.

Given: $P(X \cup Y) = \frac{11}{15}$ and $P(X) = \frac{1}{3}$.

a) Find $P(Y)$ if events $X$ and $Y$ are mutually exclusive.

Step 1: State the formula for mutually exclusive events. For mutually exclusive events, the probability of their union is the sum of their individual probabilities, as $P(X \cap Y) = 0$. $P(X \cup Y) = P(X) + P(Y)$

Step 2: Substitute the given values into the formula. $\frac{11}{15} = \frac{1}{3} + P(Y)$

Step 3: Solve for $P(Y)$. $P(Y) = \frac{11}{15} - \frac{1}{3}$ To subtract, find a common denominator, which is 15. $P(Y) = \frac{11}{15} - \frac{1 \times 5}{3 \times 5}$ $P(Y) = \frac{11}{15} - \frac{5}{15}$ $P(Y) = \frac{11 - 5}{15}$ $P(Y) = \frac{6}{15}$ Simplify the fraction: $P(Y) = \frac{2}{5}$ The probability $P(Y)$ for mutually exclusive events is $\boxed{\frac{2}{5}}$.

b) Find $P(Y)$ if events $X$ and $Y$ are independent.

Step 1: State the formula for independent events. For independent events, the probability of their intersection is the product of their individual probabilities: $P(X \cap Y) = P(X) \cdot P(Y)$. The general addition rule for probabilities is: $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$

Step 2: Substitute the independence condition into the general addition rule. $P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)$

Step 3: Substitute the given values into the equation. $\frac{11}{15} = \frac{1}{3} + P(Y) - \left(\frac{1}{3}\right)P(Y)$

Step 4: Rearrange the equation to solve for $P(Y)$. Combine terms involving $P(Y)$: $\frac{11}{15} - \frac{1}{3} = P(Y) - \frac{1}{3}P(Y)$ $\frac{11}{15} - \frac{5}{15} = P(Y)\left(1 - \frac{1}{3}\right)$ $\frac{6}{15} = P(Y)\left(\frac{3}{3} - \frac{1}{3}\right)$ $\frac{2}{5} = P(Y)\left(\frac{2}{3}\right)$

Step 5: Isolate $P(Y)$. $P(Y) = \frac{2}{5} \div \frac{2}{3}$ $P(Y) = \frac{2}{5} \times \frac{3}{2}$ $P(Y) = \frac{3}{5}$ The probability $P(Y)$ for independent events is $\boxed{\frac{3}{5}}$.

What's next?