Complementary Events in Probability
In probability, every event has a complement — the event that it does not occur. If E is an event, then the complement of E (written as E′ or not E) represents all outcomes where E does not happen.
The most important property of complementary events is that their probabilities always add up to 1. This gives a powerful shortcut: if it is easier to calculate the probability of an event not happening, subtract from 1.
Complementary events are used extensively in Class 9 and Class 10 probability problems.
What is Complementary Events in Probability?
Definition: Two events E and E′ are complementary if:
- E′ occurs whenever E does not occur, and vice versa.
- E and E′ together cover all possible outcomes.
- E and E′ cannot occur simultaneously.
P(E) + P(E′) = 1
Equivalently:
- P(E′) = 1 − P(E)
- P(E) = 1 − P(E′)
Complementary Events in Probability Formula
Key Formulas:
1. Complement Rule:
P(E) + P(not E) = 1
2. Finding complement probability:
- P(not E) = 1 − P(E)
3. Sum of all outcome probabilities = 1
Derivation and Proof
Why P(E) + P(E′) = 1:
- Let total trials = n, trials where E occurs = m.
- Trials where E does not occur = n − m.
- P(E) = m/n
- P(E′) = (n − m)/n = 1 − m/n = 1 − P(E)
- Therefore: P(E) + P(E′) = 1.
When to Use the Complement:
- When P(E) is hard to calculate directly but P(E′) is easy.
- Example: P(at least one head in 3 tosses) = 1 − P(no heads).
Types and Properties
Examples of Complementary Pairs:
1. Coin Toss: E = heads, E′ = tails.
2. Die Roll: E = even number, E′ = odd number.
3. Quality Check: E = defective, E′ = non-defective.
4. Exam: E = pass, E′ = fail.
Important: "Getting 1" and "getting 6" on a die are NOT complementary — they do not cover all outcomes.
Solved Examples
Example 1: Example 1: Basic complement
Problem: P(rain) = 0.35. Find P(no rain).
Solution:
- P(no rain) = 1 − 0.35 = 0.65
Answer: P(no rain) = 0.65.
Example 2: Example 2: Defective items
Problem: P(defective pen) = 0.08 from a batch of 500. Find P(non-defective).
Solution:
- P(non-defective) = 1 − 0.08 = 0.92
Answer: P(non-defective) = 0.92.
Example 3: Example 3: Die rolling
Problem: A die is thrown. Find P(not getting 4).
Solution:
- P(4) = 1/6
- P(not 4) = 1 − 1/6 = 5/6
Answer: P(not 4) = 5/6.
Example 4: Example 4: Prime numbers on a die
Problem: A die is thrown 100 times: 1(14), 2(18), 3(16), 4(20), 5(15), 6(17). Find P(not prime).
Solution:
- Primes: 2, 3, 5. P(prime) = (18+16+15)/100 = 0.49
- P(not prime) = 1 − 0.49 = 0.51
Answer: P(not prime) = 0.51.
Example 5: Example 5: Finding event from complement
Problem: P(student fails) = 0.15. Find P(student passes).
Solution:
- P(pass) = 1 − 0.15 = 0.85
Answer: P(pass) = 0.85.
Example 6: Example 6: Sum of all probabilities
Problem: P(A) = 0.3, P(B) = 0.45. If A, B, C are the only outcomes, find P(C).
Solution:
- P(C) = 1 − 0.3 − 0.45 = 0.25
Answer: P(C) = 0.25.
Example 7: Example 7: Bag of marbles
Problem: In 50 draws: red = 18, blue = 20. Find P(green).
Solution:
- P(green) = 1 − 18/50 − 20/50 = 12/50 = 0.24
Answer: P(green) = 0.24.
Example 8: Example 8: Verifying complementary events
Problem: From 200 numbered cards: even drawn 106 times. Verify P(even) + P(odd) = 1.
Solution:
- P(even) = 106/200 = 0.53
- P(odd) = 94/200 = 0.47
- 0.53 + 0.47 = 1 ✓
Answer: Verified — they are complementary.
Real-World Applications
Applications:
- Risk Assessment: P(machine works) = 1 − P(failure).
- Quality Control: P(non-defective) = 1 − P(defective).
- Weather Planning: P(no rain) = 1 − P(rain).
- "At Least" Problems: P(at least one success) = 1 − P(no success).
Key Points to Remember
- Complementary events cover all outcomes and are mutually exclusive.
- P(E) + P(E′) = 1.
- P(E′) = 1 − P(E).
- Useful when P(E) is hard to compute but P(E′) is easy.
- Not all event pairs are complementary — only those that exhaust all outcomes.
- Sum of all outcome probabilities = 1.
- Complementary events come in pairs: E and not-E.
Practice Problems
- P(red ball) = 0.4. Find P(not red).
- A die is thrown. Find P(not getting a number > 4).
- P(left-handed) = 0.12 in a class of 50. Find P(right-handed).
- P(A) = 0.2, P(B) = 0.35, P(C) = 0.15. Find P(D) if A, B, C, D are all outcomes.
- Heads appeared 162 times in 300 tosses. Find P(tails) and verify complement rule.
- Event E occurred 52 times in 80 trials. Find P(E) and P(not E). Verify sum = 1.
Frequently Asked Questions
Q1. What are complementary events?
Two events are complementary if one occurs exactly when the other does not. Their probabilities sum to 1.
Q2. What is the complement formula?
P(E) + P(not E) = 1. So P(not E) = 1 − P(E).
Q3. Are 'getting 1' and 'getting 6' complementary?
No. They do not cover all outcomes (2, 3, 4, 5 are missing). Complementary events must cover every possible outcome.
Q4. When is the complement rule useful?
When P(opposite event) is easier to calculate. For example, P(at least one head) = 1 − P(all tails).
Q5. Can the complement have probability 0?
Yes. If P(E) = 1 (certain), then P(not E) = 0 (impossible).
Q6. Is this in the CBSE syllabus?
Yes. Complementary events are in Chapter 15 (Probability) of NCERT Class 9 and Class 10.
Q7. Difference between complementary and mutually exclusive?
Mutually exclusive events cannot occur together. Complementary events are mutually exclusive AND exhaustive (cover all outcomes).
Q8. Can there be more than two complementary events?
No. Complementary events always come in pairs: E and not-E.
