If P(A)=0.4 , P(B)=0.6 and P(A∩B)=0.15, then the value of P(A∣A ′∪B ′ ) is

Explanation

To find \( P(A | A' \cup B') \), we can use the formula for conditional probability:

\[ P(A | A' \cup B') = \frac{P(A \cap (A' \cup B'))}{P(A' \cup B')} \]

### Step 1: Calculate \( P(A' \cup B') \)

Using the formula for the probability of the union of two events:

\[ P(A' \cup B') = P(A') + P(B') - P(A' \cap B') \]

We know that:

- \( P(A') = 1 - P(A) = 1 - 0.4 = 0.6 \)
- \( P(B') = 1 - P(B) = 1 - 0.6 = 0.4 \)

Next, we need to find \( P(A' \cap B') \). Using De Morgan's law:

\[ P(A' \cap B') = 1 - P(A \cup B) \]

To find \( P(A \cup B) \), we use the formula:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substituting the known values:

\[ P(A \cup B) = 0.4 + 0.6 - 0.15 = 0.85 \]

Now we can find \( P(A' \cap B') \):

\[ P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.85 = 0.15 \]

Now substituting back into the equation for \( P(A' \cup B') \):

\[ P(A' \cup B') = P(A') + P(B') - P(A' \cap B') = 0.6 + 0.4 - 0.15 = 0.85 \]

### Step 2: Calculate \( P(A \cap (A' \cup B')) \)

Using the distributive property:

\[ P(A \cap (A' \cup B')) = P(A \cap A') + P(A \cap B') \]

Since \( P(A \cap A') = 0 \) (an event and its complement cannot occur together), we have:

\[ P(A \cap (A' \cup B')) = 0 + P(A \cap B') = P(A) - P(A \cap B) \]

Now, we need \( P(A \cap B') \):

\[ P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.15 = 0.25 \]

### Step 3: Substitute into the conditional probability formula

Now we can substitute back into the formula for conditional probability:

\[ P(A | A' \cup B') = \frac{P(A \cap (A' \cup B'))}{P(A' \cup B')} = \frac{0.25}{0.85} \]

### Step 4: Simplify the fraction

To simplify \( \frac{0.25}{0.85} \):

\[ \frac{0.25}{0.85} = \frac{25}{85} = \frac{5}{17} \]

Thus, the final answer is:

\[ P(A | A' \cup B') = \frac{5}{17} \]