Complementary Events: Probability Relationship

In probability, complementary events are two events that together cover all possible outcomes of an experiment, and they do not overlap (i.e., they are mutually exclusive). If we know the probability of an event occurring, we can easily find the probability of that event not occurring, and vice-versa. This is because the sum of the probabilities of an event and its complement must equal 1 (or 100%).

Formulae

The fundamental relationship for complementary events is:

$P(E) + P(\text{not } E) = 1$

Where:

• $P(E)$ represents the probability of event E occurring.
• $P(\text{not } E)$ represents the probability of event E not occurring (also denoted as $P(E’)$ or $P(\overline{E})$).

We can also rewrite the formula to find the probability of the complement:

$P(\text{not } E) = 1 – P(E)$

Examples

Example-1: Suppose you roll a fair six-sided die. Let event E be rolling a ‘2’. Then:

• $P(E) = \frac{1}{6}$ (There is one ‘2’ on a six-sided die).
• $\text{not } E$ is rolling any number *other* than a ‘2’.
• $P(\text{not } E) = \frac{5}{6}$ (There are five other numbers).
• Therefore, $P(E) + P(\text{not } E) = \frac{1}{6} + \frac{5}{6} = 1$.

Example-2: A bag contains 10 marbles: 4 red and 6 blue. Let event E be selecting a red marble.

• $P(E) = \frac{4}{10} = \frac{2}{5}$
• $\text{not } E$ is selecting a blue marble.
• $P(\text{not } E) = \frac{6}{10} = \frac{3}{5}$
• Therefore, $P(E) + P(\text{not } E) = \frac{2}{5} + \frac{3}{5} = 1$.

Theorem with Proof

Theorem: For any event E in a sample space S, the sum of the probability of E and the probability of its complement (not E) is equal to 1.

Proof:

1. Let S be the sample space (the set of all possible outcomes). Then, the total probability of all outcomes in S is 1: $P(S) = 1$.
2. Event E and its complement (not E), denoted as E’, cover the entire sample space S. Every outcome is either in E or in E’ (but not both).
3. Since E and E’ are mutually exclusive (they don’t overlap), we can write: $S = E \cup E’$.
4. Therefore, the probability of S is equal to the sum of the probabilities of E and E’: $P(S) = P(E) + P(E’)$.
5. We know that $P(S) = 1$. Substituting this into the previous equation, we get: $1 = P(E) + P(E’)$.
6. Hence, $P(E) + P(\text{not } E) = 1$. Q.E.D.

Common mistakes by students

Common mistakes when dealing with complementary events include:

• Confusing the Event with its Complement: Students sometimes mistakenly identify the event instead of its complement (e.g., thinking “rolling an even number” is the complement of “rolling a 1”).
• Incorrectly Calculating Probability: Students might miscalculate the probability of an event or its complement (e.g., miscounting favorable outcomes).
• Forgetting the Total Probability Rule: Students might struggle to apply the $P(E) + P(\text{not } E) = 1$ rule, especially when the problem isn’t explicitly stated in terms of an event and its complement.

Real Life Application

Complementary events have many real-life applications, including:

• Insurance: Calculating the probability of an event (e.g., a car accident) allows insurance companies to assess the probability of the opposite outcome (e.g., no accident) to determine premiums.
• Medical Testing: Evaluating the probability of a disease based on a test result, taking into account the false positive and false negative rates, uses complementary event logic.
• Quality Control: In manufacturing, knowing the probability of a defective product helps determine the probability of a non-defective product, aiding in process improvement.
• Sports Analytics: Analyzing the probability of a team winning versus losing.

Fun Fact

The concept of complementary events is fundamental to many areas of probability and statistics. It’s a building block for understanding more complex probability distributions and statistical inference techniques.

Recommended YouTube Videos for Deeper Understanding

Q.1 If the probability of rain tomorrow is 0.3, what is the probability that it will not rain tomorrow?

Check Solution

Ans: B

$P(\text{not rain}) = 1 – P(\text{rain}) = 1 – 0.3 = 0.7$

Q.2 A bag contains only red and blue marbles. The probability of selecting a red marble is $\frac{2}{5}$. What is the probability of not selecting a red marble?

Check Solution

Ans: B

$P(\text{not red}) = 1 – P(\text{red}) = 1 – \frac{2}{5} = \frac{3}{5}$

Q.3 The probability of a student passing a math test is 0.65. What is the probability of the student failing the test?

Check Solution

Ans: A

$P(\text{fail}) = 1 – P(\text{pass}) = 1 – 0.65 = 0.35$

Q.4 A fair six-sided die is rolled. What is the probability of not rolling a 4?

Check Solution

Ans: B

$P(\text{not 4}) = 1 – P(4) = 1 – \frac{1}{6} = \frac{5}{6}$

Q.5 In a survey, 80% of people said they liked chocolate ice cream. What percentage of people did not like chocolate ice cream?

Check Solution

Ans: B

$P(\text{not like}) = 100\% – P(\text{like}) = 100\% – 80\% = 20\%$

Next Topic: Probability Problems: Single Events