Question

Given any two events A and B are such that \( P(A) = \frac{1}{2}, P(B) = \frac{1}{4} \) and \( P(A \cap B) = \frac{1}{8} \), then find \( P(\text{not A and not B}) \).

Hint:

De Morgan's laws (\( (A \cup B)' = A' \cap B' \) and \( (A \cap B)' = A' \cup B' \)) are extremely useful for transforming probability problems involving 'and' and 'or' with complements.

Answer 1

Step 1: Understanding the Concept:
We need to find the probability of the event "not A and not B", which is denoted as \( P(A' \cap B') \). We can find this using De Morgan's laws and the rule for the probability of a complement.
Step 2: Key Formula or Approach:
1. De Morgan's Law: \( A' \cap B' = (A \cup B)' \).
2. Probability of a complement: \( P(E') = 1 - P(E) \).
3. Addition rule for probability: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
Step 3: Detailed Explanation or Calculation:
We want to find \( P(A' \cap B') \).
Using De Morgan's Law, we can write this as:
\[ P(A' \cap B') = P((A \cup B)') \] Using the rule for complementary events:
\[ P((A \cup B)') = 1 - P(A \cup B) \] Now, we need to find \( P(A \cup B) \) using the addition rule:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute the given values:
\[ P(A \cup B) = \frac{1}{2} + \frac{1}{4} - \frac{1}{8} \] To add these fractions, find a common denominator, which is 8:
\[ P(A \cup B) = \frac{4}{8} + \frac{2}{8} - \frac{1}{8} = \frac{4+2-1}{8} = \frac{5}{8} \] Finally, substitute this value back to find \( P(A' \cap B') \):
\[ P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{5}{8} = \frac{3}{8} \] Step 4: Final Answer:
The probability of P(not A and not B) is \( \frac{3}{8} \).