Step 1: Recall the formula for conditional probability
The conditional probability \( P(F|E) \) is given by: \[ P(F|E) = \frac{P(E \cap F)}{P(E)}. \] Step 2: Find \( P(E \cap F) \)
Using the formula for the probability of the union of two events: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F). \] Substitute the given values \( P(E \cup F) = 0.4 \), \( P(E) = 0.1 \), \( P(F) = 0.3 \): \[ 0.4 = 0.1 + 0.3 - P(E \cap F). \] Simplify to find \( P(E \cap F) \): \[ P(E \cap F) = 0.1 + 0.3 - 0.4 = 0. \] Step 3: Calculate \( P(F|E) \)
Substitute \( P(E \cap F) = 0 \) and \( P(E) = 0.1 \) into the formula for \( P(F|E) \): \[ P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{0}{0.1} = 0. \] Step 4: Conclude the result
The conditional probability \( P(F|E) \) is \( 0 \). This indicates that the events \( E \) and \( F \) do not overlap.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find:
\(Evaluate \ P(A∩B)\ if \ 2P(A) = P(B) =\) \(\frac {5}{13}\) \(and \ P(A|B)=\) \(\frac 25\)