Two events E and F are independent. If P(E )=0.3 and P(`EcupF`)=0.5 then `P(E//F)-P(F//E)` equals to

Options

A. `2/7`
B. `3/35`
C. `1/70`
D. `1/7`

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understanding the Given Information
We are given:
- \( P(E) = 0.3 \)
- \( P(E \cup F) = 0.5 \)

We need to find \( P(E|F) - P(F|E) \).

### Step 2: Using the Formula for Union of Two Events
We know that for any two events \( E \) and \( F \):
\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
Since \( E \) and \( F \) are independent, we have:
\[ P(E \cap F) = P(E) \cdot P(F) \]

### Step 3: Setting Up the Equation
Let \( P(F) = x \). Then, substituting the known values into the union formula:
\[ 0.5 = 0.3 + x - (0.3 \cdot x) \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ 0.5 = 0.3 + x - 0.3x \]
This simplifies to:
\[ 0.5 = 0.3 + x(1 - 0.3) \]
\[ 0.5 = 0.3 + 0.7x \]

### Step 5: Solving for \( x \)
Now, isolate \( x \):
\[ 0.5 - 0.3 = 0.7x \]
\[ 0.2 = 0.7x \]
\[ x = \frac{0.2}{0.7} = \frac{2}{7} \]
Thus, \( P(F) = \frac{2}{7} \).

### Step 6: Finding Conditional Probabilities
Now we can find \( P(E|F) \) and \( P(F|E) \):
- \( P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{P(E) \cdot P(F)}{P(F)} = P(E) = 0.3 \)
- \( P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{P(E) \cdot P(F)}{P(E)} = P(F) = \frac{2}{7} \)

### Step 7: Calculating the Final Result
Now we can calculate:
\[ P(E|F) - P(F|E) = 0.3 - \frac{2}{7} \]
Converting \( 0.3 \) to a fraction:
\[ 0.3 = \frac{3}{10} \]
Now we need a common denominator to subtract:
- The least common multiple of 10 and 7 is 70.

Converting both fractions:
\[ \frac{3}{10} = \frac{21}{70} \]
\[ \frac{2}{7} = \frac{20}{70} \]
Now subtracting:
\[ P(E|F) - P(F|E) = \frac{21}{70} - \frac{20}{70} = \frac{1}{70} \]

### Final Answer
Thus, the final answer is:
\[ P(E|F) - P(F|E) = \frac{1}{70} \]

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