If A and B are any two events such that $$P(A) = \frac{2}{5}$$ and $$P(A \cap B) = \frac{3}{20}$$, then the conditional probability, $$P(A|(A' \cup B'))$$, where $$A'$$ denotes the complement of A, is equal to:

We are given that $$P(A) = \frac{2}{5}$$ and $$P(A \cap B) = \frac{3}{20}$$. We need to find $$P(A \mid (A' \cup B'))$$, where $$A'$$ and $$B'$$ are the complements of events $$A$$ and $$B$$ respectively.

The conditional probability is defined as $$P(A \mid C) = \frac{P(A \cap C)}{P(C)}$$, where $$C = A' \cup B'$$. So, we have:

$$P(A \mid (A' \cup B')) = \frac{P(A \cap (A' \cup B'))}{P(A' \cup B')}$$

First, simplify the numerator $$P(A \cap (A' \cup B'))$$. Using the distributive law of sets:

$$A \cap (A' \cup B') = (A \cap A') \cup (A \cap B')$$

Since $$A \cap A' = \emptyset$$ (the empty set, as $$A$$ and its complement are disjoint), this simplifies to:

$$A \cap (A' \cup B') = \emptyset \cup (A \cap B') = A \cap B'$$

Thus, $$P(A \cap (A' \cup B')) = P(A \cap B')$$.

Now, $$A \cap B'$$ represents the event that $$A$$ occurs but $$B$$ does not. This can also be written as $$A \setminus (A \cap B)$$. Since $$A \cap B \subseteq A$$, we have:

$$P(A \cap B') = P(A) - P(A \cap B)$$

Substitute the given probabilities:

$$P(A) = \frac{2}{5} = \frac{8}{20}, \quad P(A \cap B) = \frac{3}{20}$$

So,

$$P(A \cap B') = \frac{8}{20} - \frac{3}{20} = \frac{5}{20} = \frac{1}{4}$$

Therefore, the numerator is $$\frac{1}{4}$$.

Next, simplify the denominator $$P(A' \cup B')$$. Using De Morgan's law:

$$A' \cup B' = (A \cap B)'$$

So,

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

Substitute $$P(A \cap B) = \frac{3}{20}$$:

$$P(A' \cup B') = 1 - \frac{3}{20} = \frac{20}{20} - \frac{3}{20} = \frac{17}{20}$$

Therefore, the denominator is $$\frac{17}{20}$$.

Now, the conditional probability is:

$$P(A \mid (A' \cup B')) = \frac{\frac{1}{4}}{\frac{17}{20}}$$

Dividing fractions is equivalent to multiplying by the reciprocal:

$$\frac{\frac{1}{4}}{\frac{17}{20}} = \frac{1}{4} \times \frac{20}{17} = \frac{1 \times 20}{4 \times 17} = \frac{20}{68}$$

Simplify $$\frac{20}{68}$$ by dividing both numerator and denominator by 4:

$$\frac{20 \div 4}{68 \div 4} = \frac{5}{17}$$

Thus, $$P(A \mid (A' \cup B')) = \frac{5}{17}$$.

Comparing with the options:

• A. $$\frac{11}{20}$$
• B. $$\frac{5}{17}$$
• C. $$\frac{8}{17}$$
• D. $$\frac{1}{4}$$

Hence, the correct answer is Option B.