Complementary Events

Complementary events are pairs of outcomes in probability where one event occurs if and only if the other does not. They arise in situations where there are exactly two outcomes. The probabilities of complementary events always add up to 1. Understanding this concept is useful for simplifying probability calculations. In this guide, you will learn about complementary events, its definition, formula, and related examples, helping you solve problems more quickly and confidently.

Table of Contents

• What are Complementary Events
• Properties of Complementary Events
• Solved Examples on Complementary Events

What are Complementary Events

For an experiment with sample space S and event E, there exists another event Ec that contains the remaining elements of the sample space. Ec is called the complementary event of E.

If E is an event. Its complement is denoted as Ec or E′.

E and E′ are complementary events if:

• E′ occurs whenever E does not occur, and vice versa. Two events are said to be complementary events if one event can take place only when the other does not occur.
• E and E′ together cover all possible outcomes. i.e, Probabilities of complementary events sum to 1
• P(E) + P(E′) = 1
• P(E) = 1 - P(E′)
• E and E′ cannot occur simultaneously.

Properties of Complementary Events

For two events to be considered complementary, they must satisfy certain important properties. These are explained below:

• Complementary events are mutually exclusive, which means they cannot occur at the same time. If one event happens, the other cannot happen. In other words, they are disjoint events with no common outcomes.  A∩A′=∅
• Complementary events are exhaustive. They cover the entire sample space. This implies that either the event or its complement must occur. Mathematically, this can be written as: S = A \cup A' where S is the sample space, A is the event, and A' is its complement.

Solved Examples on Complementary Events

Example 1: A random number is chosen from 1 to 50. Calculate the probability of not choosing a perfect square.

Solution: Let E be the event of choosing a perfect number. Total number of outcomes = 50

S = {1, 4, 9, 16 , 25 , 36 , 49}

P(E) = 7/50 

Probability of not choosing a perfect square = P(E′) = 1 - P(E) = 1 - (7/50) = 43/50 = 0.86

Example 2: From a batch of 500, P(defective pen) = 0.07. Find P(non-defective).

Solution: P(non-defective) = 1 - P(defective pen) (∵ the events are complementary)

P(non-defective) = 1 - 0.07 = 0.93.

Example 3: A die is thrown. Find the probability of not getting 4.

Solution: Total outcome = 6

S = {1,2,3,4,5,6} 

Let E be the event of getting a 4.

P(E) = 1/6.

The probability of not getting 4 = P (E′) = 1 - P(E) = 1 - (1/6) = (5/6) = 0.833.