# Complimentary events in probability

In this chapter, we will discuss the concept of complimentary events in probability with examples.

At the end of the chapter, solved problems are given for your understanding.

## What are complimentary events ?

Two events A and negation of A are complimentary in nature if occurrence of one event prevents the occurrence of other event.

For example;
Consider the experiment of tossing a coin.

The two possible outcomes are: Heads or Tails.

Sample space (S) = { Head, Tail }

Let A be the event of getting head and B be the event of getting tail.

We know that if event A occurs then B can’t happen and vice-versa. So we can say that A and B are complimentary events.

Note:
On combining both the complimentary events, we will get the sample space back.

### Representing complimentary events

If two events A & B are complimentary, then mathematically we can also express the events as;

B = \mathtt{A'\ \ or\ \ \overline{A} \ \ or\ neg( A)}

It tells that event B contain element that is not in event A.

### Probability of complementary events

If A & B are complimentary events then the sum of their probability is 1.

P(A) + P(B) = 1

Let us prove the above expression with example.

Consider the same above experiment of tossing an unbiased coin.
Sample Space = { Head, Tail }

Where;
Event A = getting head
Event B = getting tail

Both event A & B are complimentary.
P (A) = 1/2 = 0.5
P (B) = 1/2 = 0.5

Adding the probability of both events.
P (A) + P(B) = 1

Hence, on adding probability of complimentary events we will get 1.

### Complimentary events – Solved problems

Question 01
Consider the experiment of throwing a dice. A is the event of getting odd number and B is the event of getting even number. Find the probability of event A & B and check if they are complimentary event.

Solution
When a dice is thrown, you can get number from 1 to 6.
Sample Space (S) = {1, 2, 3, 4, 5, 6 }

Event A = Outcome is odd number.
A = { 1, 3, 5 }

Event B = Outcome is even number
B = { 2, 4, 6 }

Note that there is no common element between event A & B. It mean that if event A occur then B doesn’t occur and vice-versa.

Hence, both event A & B are complimentary in nature.

Now calculating the probability;
P (A) = 3/6 = 0.5
P (B) = 3/6 = 0.5

Adding both the probability, we will get 1.

Question 02
John is waiting on a station for bus to arrive. The correct time for arrival of bus of 5 : 00 PM. The probability that the bus will arrive on time is 3/4. What is the probability that bus will run late ?

Solution
For the above experiment, there are two possible outcomes, bus arrived on time or bus is late.

Sample Space (S) = { ” on time”, “running late ” }

Let A be the event for bus running late.
A = { “running late ” }

And B be the event for bus arriving on time.
B ={ “on time” }

Note that both events A and B are complimentary since occurrence of one event mean negation of other event.

We know that sum of probability of complimentary event is 1.

P(A) + P(B) = 1

P(A) + 3/4 = 1

P(A) = 1- 3/4

P(A) = 1/4

Hence, probability of bus running late is 1/4.

Question 03
In an experiment, a dice is rolled once. Consider the probability of getting number divisible by 3.

Solution
Given below is the sample space;
S = {1, 2, 3, 4, 5, 6}

So there are 6 possible outcome.

Let A be the event of getting number divisible by 3.
A = { 3, 6 }

There are 2 favorable outcomes.

Calculating probability for event A;
P(A) = 2 / 6 = 1/3

Hence, the required probability is 1/3.

Question 04
In a box of 200 apples, there are 7 rotten ones. One apple is picked randomly from the box. Find the probability of getting fresh apples.

Solution
Let A be the event of getting rotten apples and B is the event of getting fresh apples.

Note that both the events A & B are complimentary since if event A occur then event B can’t happen.

Calculating probability of rotten apples, P(A);
P (A) = 7 / 200

For complimentary events;
P(A) + P(B) = 1

7/200 + P(B) = 1

P(B) = 1 – 7/200

P(B) = 193/200

Hence, probability of getting fresh apple is 193/200

Question 05
In a box, there are 6 tennis balls and rest are cricket balls. One ball is selected at random from the box. If probability of getting tennis ball is 0.75 then find the probability of selecting cricket ball.

Solution
Let A be the event of selecting tennis ball and B be the event of getting cricket ball.

Understand that both the events A & B are complimentary because if event A occur then B will not occur.

In case of complimentary events;
P(A) + P(B) = 1

0.75 + P(B) = 1

P(B) = 0.25

Hence, probability of selecting cricket ball is 0.25

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