Math Problem Statement

Explain the term independent events. Also show that, if A and B are independent events then i) ii) AC and B are also independent. AC and BC are also independent.

Solution

Independent Events

Two events AA and BB are called independent if the occurrence of one event does not affect the probability of the other event occurring. Mathematically, events AA and BB are independent if:

P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

This means that the probability of both events happening together is equal to the product of their individual probabilities.

Proof of Independence of Complementary Events

1. ACA^C and BB are Independent

Let ACA^C denote the complement of event AA, meaning that ACA^C represents the event that AA does not occur. We want to show that ACA^C and BB are independent, given that AA and BB are independent.

We know from the definition of independent events that:

P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

Now, the complement of AA, ACA^C, is everything except AA, so:

P(ACB)=P(B)P(AB)P(A^C \cap B) = P(B) - P(A \cap B)

Substitute P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) into the equation:

P(ACB)=P(B)P(A)×P(B)P(A^C \cap B) = P(B) - P(A) \times P(B)

Factor out P(B)P(B):

P(ACB)=P(B)×(1P(A))P(A^C \cap B) = P(B) \times (1 - P(A))

Since 1P(A)=P(AC)1 - P(A) = P(A^C), we have:

P(ACB)=P(AC)×P(B)P(A^C \cap B) = P(A^C) \times P(B)

This shows that ACA^C and BB are independent.

2. ACA^C and BCB^C are Independent

Next, we show that ACA^C and BCB^C, the complements of events AA and BB, are also independent.

We know that:

P(ACBC)=1P(AB)P(A^C \cap B^C) = 1 - P(A \cup B)

Using the inclusion-exclusion principle, we can write:

P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Since P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) (due to the independence of AA and BB), we substitute this into the equation:

P(ACBC)=1(P(A)+P(B)P(A)×P(B))P(A^C \cap B^C) = 1 - (P(A) + P(B) - P(A) \times P(B))

Simplify:

P(ACBC)=1P(A)P(B)+P(A)×P(B)P(A^C \cap B^C) = 1 - P(A) - P(B) + P(A) \times P(B)

This can be factored as:

P(ACBC)=(1P(A))×(1P(B))=P(AC)×P(BC)P(A^C \cap B^C) = (1 - P(A)) \times (1 - P(B)) = P(A^C) \times P(B^C)

Thus, ACA^C and BCB^C are independent.


Conclusion

We have shown that:

  • If AA and BB are independent events, then:
    1. ACA^C and BB are also independent.
    2. ACA^C and BCB^C are also independent.

Would you like further clarification on any part of this proof?

Here are some related questions for deeper understanding:

  1. What are dependent events, and how do they differ from independent events?
  2. How does the inclusion-exclusion principle apply in probability?
  3. Can two mutually exclusive events be independent? Why or why not?
  4. How does conditional probability relate to independent events?
  5. What are some real-life examples of independent events?

Tip: When solving probability problems, always check whether events are independent before applying the multiplication rule for combined probabilities.

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Math Problem Analysis

Mathematical Concepts

Probability
Independent Events
Complementary Events

Formulas

P(A ∩ B) = P(A) × P(B)
P(A^C ∩ B) = P(B) - P(A) × P(B)
P(A^C ∩ B^C) = (1 - P(A)) × (1 - P(B))

Theorems

Independence of Events Theorem
Inclusion-Exclusion Principle

Suitable Grade Level

Grades 10-12