If P(A) = 0 . 4 , P(B')= 0.6 and `P(A cap B)` = 0 . 15, then the value of `P(A//A'cup B')` is

Options

A. `(1)/(17)`
B. `(4)/(17)`
C. `(5)/(17)`
D. `(10)/(17)`

Answer

To find the value of \( P(A | A' \cup B') \), 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 union of two events:

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

We know \( P(A') = 1 - P(A) = 1 - 0.4 = 0.6 \) and \( P(B') = 0.6 \).

Next, we need to find \( P(A' \cap B') \). We can use the complement of the intersection of events:

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

Using the formula for the union of two events:

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

We have \( P(A) = 0.4 \) and \( P(A \cap B) = 0.15 \). To find \( P(B) \), we can use the relationship \( P(B') = 1 - P(B) \):

\[ P(B) = 1 - P(B') = 1 - 0.6 = 0.4 \]

Now substituting into the union formula:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.4 - 0.15 = 0.65 \]

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

\[ P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.65 = 0.35 \]

Now substituting back to find \( P(A' \cup B') \):

\[ P(A' \cup B') = P(A') + P(B') - P(A' \cap B') = 0.6 + 0.6 - 0.35 = 0.85 \]

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

Since \( A \cap (A' \cup B') = (A \cap A') \cup (A \cap B') \), and \( A \cap A' \) is empty (probability is 0), we have:

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

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

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

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

Now we can substitute back into the conditional probability formula:

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

### 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 value of \( P(A | A' \cup B') \) is:

\[ \boxed{\frac{5}{17}} \]