The given events \( A \) and \( B \) are such that \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{2} \), and \( P(A \cap B) = \frac{1}{8} \); then find \( P(A' \cap B') \).
Step 1: Using the formula for the probability of the complement: \[ P(A \text{ not} \cap B \text{ not}) = 1 - P(A \cup B). \] Step 2: Use the inclusion-exclusion principle to calculate \( P(A \cup B) \): \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \] Substitute the given values: \[ P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{5}{8}. \] Step 3: Now calculate the complement: \[ P(A \text{ not} \cap B \text{ not}) = 1 - \frac{5}{8} = \frac{3}{8}. \] Thus, the answer is \( \frac{3}{8} \).
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $