Show that if events A B and C are independent that the following pairs of events are also independent:
a) A and B Union C
b) A - B and C
a) To show that events A, B, and C being independent implies that A and B Union C are also independent, we need to prove that the probability of the intersection of A and B Union C is equal to the product of their individual probabilities.
Let's start by expanding A and B Union C:
A and B Union C = (A and B) Union (A and C)
We can rewrite the probability of A and B Union C as follows:
P(A and B Union C) = P((A and B) Union (A and C))
Using the principle of inclusion-exclusion, the probability of the union of two events can be written as the sum of their individual probabilities minus the probability of their intersection:
P((A and B) Union (A and C)) = P(A and B) + P(A and C) - P(A and B and C)
Since A, B, and C are independent events, we know that:
P(A and B) = P(A) * P(B)
P(A and C) = P(A) * P(C)
P(A and B and C) = P(A) * P(B) * P(C)
Substituting these values back into our expression:
P((A and B) Union (A and C)) = P(A) * P(B) + P(A) * P(C) - P(A) * P(B) * P(C)
Now, let's consider the product of the individual probabilities of A and B Union C:
P(A) * P(B Union C) = P(A) * [P(B) + P(C) - P(B) * P(C)]
We can see that the two expressions are equal:
P((A and B) Union (A and C)) = P(A) * P(B) + P(A) * P(C) - P(A) * P(B) * P(C) = P(A) * [P(B) + P(C) - P(B) * P(C)] = P(A) * P(B Union C)
Therefore, we have shown that if events A, B, and C are independent, then A and B Union C are also independent.
b) To prove that events A - B and C are independent, we need to show that the probability of their intersection is equal to the product of their individual probabilities.
Let's start by expanding A - B:
A - B = A and B'
We can rewrite the probability of A - B as follows:
P(A - B) = P(A and B')
Since A and B are independent, we know that:
P(A and B) = P(A) * P(B)
Substituting this value back into our expression:
P(A - B) = P(A) * P(B')
Now, consider the product of the individual probabilities of A - B and C:
P(A - B) * P(C) = P(A) * P(B') * P(C)
To show that events A - B and C are independent, we need to prove that P(A - B) * P(C) = P(A - B and C).
Let's rewrite the probability of A - B and C as follows:
P(A - B and C) = P(A and B') * P(C)
Since A and B are independent, we know that:
P(A and B') = P(A) * P(B')
Substituting this value back into our expression:
P(A - B and C) = P(A) * P(B') * P(C)
We can see that P(A - B) * P(C) = P(A - B and C), indicating that events A - B and C are independent.
Therefore, we have shown that if events A, B, and C are independent, then A - B and C are also independent.
To show that two events are independent, we need to demonstrate that the probability of their intersection is equal to the product of their individual probabilities. In this case, we need to prove that the following pairs of events are independent:
a) A and (B Union C)
b) (A - B) and C
Before we start, let's briefly discuss some properties of set operations.
1. Union (denoted by ∪): The union of two events A and B, denoted as A ∪ B, represents the event that either A occurs, B occurs, or both events occur.
2. Difference (denoted by -): The difference between two events A and B, denoted as A - B, represents the event that A occurs, but B does not occur.
Now let's prove the independence of each pair of events:
a) A and (B Union C):
To show that A and (B Union C) are independent, we need to prove that the probability of their intersection is equal to the product of their individual probabilities.
P(A ∩ (B ∪ C)) = P(A) * P(B ∪ C)
Expanding the expression on the right side:
P(A ∩ (B ∪ C)) = P(A) * [P(B) + P(C) - P(B ∩ C)]
Since A, B, and C are independent:
P(A ∩ (B ∪ C)) = P(A) * [P(B) + P(C) - P(B) * P(C)]
Distributing P(A) over the terms inside the brackets:
P(A ∩ (B ∪ C)) = P(A) * P(B) + P(A) * P(C) - P(A) * P(B) * P(C)
Now, let's break down the left side of the equation:
P(A ∩ (B ∪ C)) represents the probability that A and (B ∪ C) both occur, which can be simplified to the probability that A, B, and C all occur.
P(A ∩ (B ∪ C)) = P(A ∩ B ∩ C)
Using the properties of independence:
P(A ∩ B ∩ C) = P(A) * P(B) * P(C)
Therefore, we have:
P(A) * [P(B) + P(C) - P(B) * P(C)] = P(A) * P(B) * P(C)
Canceling out P(A) from both sides:
P(B) + P(C) - P(B) * P(C) = P(B) * P(C)
Rearranging the terms:
P(B) + P(C) = P(B) * P(C)
This equation represents the independence of event A and (B Union C). Thus, if events A, B, and C are independent, then A and (B Union C) are also independent.
b) (A - B) and C:
Similarly, to show that (A - B) and C are independent, we need to prove that the probability of their intersection is equal to the product of their individual probabilities.
P((A - B) ∩ C) = P(A - B) * P(C)
Expanding the expression on the right side:
P((A - B) ∩ C) = (P(A) - P(A ∩ B)) * P(C)
Since A, B, and C are independent:
P((A - B) ∩ C) = (P(A) - P(A) * P(B)) * P(C)
Now, let's break down the left side of the equation:
P((A - B) ∩ C) represents the probability that (A - B) and C both occur, which can be simplified to the probability that A, B does not occur, and C occurs (i.e., A occurs and B does not occur, and C occurs).
P((A - B) ∩ C) = P(A) * (1 - P(B)) * P(C)
Using the properties of independence:
P(A) * (1 - P(B)) * P(C) = P(A) * P(C) * (1 - P(B))
Therefore, we have:
P(A) * (1 - P(B)) * P(C) = (P(A) - P(A) * P(B)) * P(C)
Canceling out P(A) * P(C) from both sides:
(1 - P(B)) = (1 - P(A) * P(B))
Rearranging the terms:
P(B) = P(A) * P(B)
This equation represents the independence of event (A - B) and C. Thus, if events A, B, and C are independent, then (A - B) and C are also independent.