Question

If A and B are two events such that P(A)=0.5, P(B)=0.4 and P(A\(\cap\)B)=0.2, then P(A|(A\(\cup\)B)) is equal to

• A. \(\frac{6}{7}\)
• B. \(\frac{5}{6}\)
• C. \(\frac{5}{7}\)
• D. \(\frac{4}{7}\)
• E. \(\frac{1}{2}\)

Answer 1

The Correct Option is C

We are given the following probabilities: \[ P(A) = 0.5, \quad P(B) = 0.4, \quad P(A \cap B) = 0.2 \] We need to find $P(A | (A \cup B))$, which is the probability of event $A$ occurring given that either $A$ or $B$ occurs. Using the conditional probability formula: \[ P(A | (A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} \] Since $A \cap (A \cup B) = A$, we have: \[ P(A | (A \cup B)) = \frac{P(A)}{P(A \cup B)} \] Now, we calculate $P(A \cup B)$ using the formula for the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the given values: \[ P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7 \] Thus: \[ P(A | (A \cup B)) = \frac{0.5}{0.7} = \frac{5}{7} \]

The correct option is (C) : \(\frac{5}{7}\)

Answer 2

We are given P(A) = 0.5, P(B) = 0.4, and \(P(A \cap B) = 0.2\). We want to find \(P(A | (A \cup B))\).

By definition of conditional probability, \(P(A | (A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}\).

First, we find \(P(A \cup B)\) using the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\):

\(P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7\)

Next, we find \(P(A \cap (A \cup B))\). Since A is a subset of \(A \cup B\), the intersection of A and \(A \cup B\) is just A. So, \(A \cap (A \cup B) = A\).

Therefore, \(P(A \cap (A \cup B)) = P(A) = 0.5\).

Now we can find the conditional probability:

\(P(A | (A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)} = \frac{0.5}{0.7} = \frac{5}{7}\)