Let X and Y be two events such that `P(X)=1/3, P(X|Y)=1/2and P(Y|X)=2/5.` Then

Options

A. `P(Y)=4/15`
B. `P(X'|Y)=1/2`
C. `P(XuuY)=2/5`
D. `P(XnnY)=1/5`

Explanation

To solve the problem step by step, we will use the given probabilities and the definitions of conditional probability and intersection of events.

### Given:
- \( P(X) = \frac{1}{3} \)
- \( P(X|Y) = \frac{1}{2} \)
- \( P(Y|X) = \frac{2}{5} \)

### Step 1: Find \( P(X \cap Y) \)
Using the definition of conditional probability, we know:
\[ P(Y|X) = \frac{P(X \cap Y)}{P(X)} \]
Substituting the known values:
\[ \frac{2}{5} = \frac{P(X \cap Y)}{\frac{1}{3}} \]
To find \( P(X \cap Y) \), we can rearrange the equation:
\[ P(X \cap Y) = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15} \]

### Step 2: Find \( P(Y) \)
Now, we can use the conditional probability \( P(X|Y) \):
\[ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} \]
Substituting the known values:
\[ \frac{1}{2} = \frac{\frac{2}{15}}{P(Y)} \]
Rearranging gives:
\[ P(Y) = \frac{2}{15} \cdot 2 = \frac{4}{15} \]

### Step 3: Find \( P(X')|Y \)
Now we need to find \( P(X'|Y) \), where \( X' \) is the complement of \( X \):
\[ P(X'|Y) = \frac{P(X' \cap Y)}{P(Y)} \]
We can express \( P(X' \cap Y) \) as:
\[ P(X' \cap Y) = P(Y) - P(X \cap Y) \]
Substituting the values we found:
\[ P(X' \cap Y) = \frac{4}{15} - \frac{2}{15} = \frac{2}{15} \]
Now substituting back into the equation for \( P(X'|Y) \):
\[ P(X'|Y) = \frac{\frac{2}{15}}{\frac{4}{15}} = \frac{2}{4} = \frac{1}{2} \]

### Step 4: Find \( P(X \cup Y) \)
Using the formula for the union of two events:
\[ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) \]
Substituting the known values:
\[ P(X \cup Y) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} \]
To add these fractions, we need a common denominator. The least common multiple of 3 and 15 is 15:
\[ P(X) = \frac{1}{3} = \frac{5}{15} \]
Now substituting:
\[ P(X \cup Y) = \frac{5}{15} + \frac{4}{15} - \frac{2}{15} = \frac{5 + 4 - 2}{15} = \frac{7}{15} \]

### Final Results:
- \( P(Y) = \frac{4}{15} \)
- \( P(X'|Y) = \frac{1}{2} \)
- \( P(X \cup Y) = \frac{7}{15} \)