Question

If x, y, z are three independent events, then prove that \(P(X \cap Y \cap Z) = P(X) \cdot P\left(\frac{Y}{X}\right) \cdot P\left(\frac{Z}{X \cap Y}\right)\).

Hint:

The multiplication rule for probability is like a chain: the probability of the first event, times the probability of the second given the first has occurred, times the probability of the third given the first two have occurred, and so on. This general rule simplifies to a simple product of probabilities, \(P(X)P(Y)P(Z)\), only in the special case where the events are independent.

Answer 1

Step 1: Understanding the Concept:
The equation to be proved is the general multiplication rule (or chain rule) for the probability of the intersection of three events. The problem states that the events are independent, which is a specific condition. We can prove the identity in two ways: by showing it holds true for any set of events, or by using the independence condition to show both sides are equal. We will prove the general rule, which is always true.
Step 2: Key Formula or Approach:
The proof relies on the definition of conditional probability: \[ P(B|A) = \frac{P(A \cap B)}{P(A)} \quad \implies \quad P(A \cap B) = P(A) \cdot P(B|A) \] Step 3: Detailed Explanation:
We want to prove \(P(X \cap Y \cap Z) = P(X) \cdot P(Y|X) \cdot P(Z|X \cap Y)\).
Let's start with the left-hand side (LHS), \(P(X \cap Y \cap Z)\). We can group the first two events and write this as: \[ P((X \cap Y) \cap Z) \] Now, apply the definition of conditional probability, treating \((X \cap Y)\) as a single event 'A' and \(Z\) as event 'B'. \[ P((X \cap Y) \cap Z) = P(X \cap Y) \cdot P(Z | (X \cap Y)) \quad \text{---(1)} \] Next, we can expand the term \(P(X \cap Y)\) using the same definition of conditional probability: \[ P(X \cap Y) = P(X) \cdot P(Y|X) \quad \text{---(2)} \] Now, substitute the expression for \(P(X \cap Y)\) from equation (2) into equation (1): \[ P(X \cap Y \cap Z) = \left[ P(X) \cdot P(Y|X) \right] \cdot P(Z | (X \cap Y)) \] \[ P(X \cap Y \cap Z) = P(X) \cdot P(Y|X) \cdot P(Z | X \cap Y) \] This is the same as the right-hand side (RHS). The identity is true for any three events, whether they are independent or not.
Note on Independence: The problem states that X, Y, and Z are independent. If this condition is applied:
By definition of independence, \(P(X \cap Y \cap Z) = P(X)P(Y)P(Z)\). (LHS)
Also by independence, \(P(Y|X) = P(Y)\) and \(P(Z|X \cap Y) = P(Z)\).
Substituting these into the RHS gives: \(P(X) \cdot P(Y) \cdot P(Z)\).
Since both sides simplify to the same expression under the condition of independence, the identity holds true for independent events as well.
Step 4: Final Answer:
By applying the definition of conditional probability twice, we have shown that the identity holds. Hence proved.