Exercise 13.1 — Complementary Events

Probability of complementary events.

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What is Probability?

Probability is the measure of how likely an event is to occur. It is one of the most important and practical topics in the Class 10 Mathematics syllabus for CBSE, Telangana, and Andhra Pradesh board exams. From predicting weather to understanding card games, probability is everywhere in real life.

In Exercise 13.1, students work with the concept of classical (theoretical) probability — calculating the chance of an event by counting favourable outcomes out of all equally likely possible outcomes.

P(E) = Number of outcomes favourable to E / Total number of all possible outcomes

0 ≤ P(E) ≤ 1 Probability always lies between 0 and 1

P(E) = 0 Impossible Event — can never happen

P(E) = 1 Sure/Certain Event — always happens

P(E) + P(Ē) = 1 Complementary Events rule

💡 Complementary Events: If E is an event, then Ē (read "not E") is the event that E does not happen. They are always complementary: the probability of one is 1 minus the probability of the other. This is used directly in Q3 and Q6.

Question 1 — Completing Key Probability Statements

This question tests whether students have understood the fundamental rules of probability. Here are the five statements explained properly:

#	Statement	Answer	Why?
(i)	P(E) + P(not E) = ?	1	Complementary events always sum to 1
(ii)	Probability of an event that cannot happen	0 — Impossible Event	No favourable outcomes exist
(iii)	Probability of an event certain to happen	1 — Sure/Certain Event	All outcomes are favourable
(iv)	Sum of probabilities of all elementary events	1	All events together cover the entire sample space
(v)	Probability of an event is ≥ ___ and ≤ ___	0 and 1	0 ≤ P(E) ≤ 1 — the fundamental bound

Question 2 — Equally Likely Outcomes: Yes or No?

Two outcomes are equally likely when there is no reason to expect one more than the other — both have an identical chance. This question asks students to think critically about real-life situations. It is not enough to say "there are two outcomes" — you must check whether those outcomes have an equal chance.

Part (i)

A driver tries to start a car. The car starts or does not start.

Answer: No — not equally likely.

Whether a car starts depends on many real-world factors: the fuel level, the battery charge, the age of the vehicle, and its maintenance history. Because these factors vary, starting and not starting do not have an equal chance. The outcomes are not equally likely.

Part (ii)

A basketball player attempts a shot. She shoots or misses.

Answer: No — not equally likely.

The outcome depends entirely on the player's skill level, which is not specified. A professional player may score 90% of the time; a beginner may miss 90% of the time. Without knowing the player's ability, we cannot say scoring and missing are equally likely. The outcomes are not equally likely.

Part (iii)

A true-false question is answered. The answer is right or wrong.

Answer: Yes — equally likely.

There are exactly two possible answers — TRUE or FALSE — and if the person guesses randomly, each has the same chance. There is no third option and no hidden bias. The outcomes are equally likely.

Part (iv)

A baby is born. It is a boy or a girl.

Answer: Yes — equally likely.

The biological probability of a newborn being male or female is approximately equal. There is no external factor that makes one more likely than the other. The outcomes are equally likely.

Questions 3 & 6 — Using the Complementary Events Rule

Problem 3

If P(E) = 0.05, find the probability of 'not E'.

Since P(E) and P(Ē) are complementary events, their probabilities must add up to 1.

P(E) + P(Ē) = 1 0.05 + P(Ē) = 1 P(Ē) = 1 − 0.05 P(Ē) = 0.95

✅ Probability of 'not E' = 0.95. This means if the event E is very unlikely (only a 5% chance), its complement is highly likely (95% chance).

Problem 6

In a group of 3 students, the probability that 2 students do NOT share a birthday is 0.992. What is the probability that they DO share the same birthday?

Let E = event that 2 students do not share a birthday. Then Ē = event that they do share a birthday. These are complementary.

P(E) = 0.992 (given) P(Ē) = 1 − P(E) P(Ē) = 1 − 0.992 P(Ē) = 0.008

The probability that the 2 students share the same birthday is 0.008 — a very small chance, which makes intuitive sense given how many days are in a year.

Question 4 — Impossible vs Certain Events (Candy Bag)

Problem 4

A bag contains only lemon-flavoured candies. Malini picks one candy without looking. What is the probability she picks (i) an orange-flavoured candy? (ii) a lemon-flavoured candy?

This problem illustrates the two extreme values of probability — 0 and 1 — using a simple everyday situation.

(i) P(orange candy) = 0/total = 0 → Impossible Event: no orange candies exist in the bag (ii) P(lemon candy) = total/total = 1 → Certain Event: every candy is lemon-flavoured

📌 Key insight: You don't need a number — you just need to reason. If the bag has only one type of candy, picking that type is certain (P = 1) and picking any other type is impossible (P = 0).

Question 5 — Playing Cards: Hearts Removed

A standard deck has 52 cards divided into 4 suits of 13 cards each. The suits and their colours are:

♥ Hearts — 13 red cards ♦ Diamonds — 13 red cards ♠ Spades — 13 black cards ♣ Clubs — 13 black cards

Each suit has one Ace, one King, one Queen, one Jack, and number cards 2 through 10 (13 cards total). There are 4 Aces in a full deck — one per suit.

Rahim removes all 13 heart cards. The remaining deck has 52 − 13 = 39 cards.

Problem 5 — Setup

Remaining deck after removing hearts: n(S) = 39

Part	Event	Favourable Outcomes	n(S)	Probability
(i)	Getting an Ace	3 (♦A, ♠A, ♣A — heart ace is removed)	39	3/39 = 1/13
(ii)	Getting a Diamond	13 (all 13 diamonds remain)	39	13/39 = 1/3
(iii)	Getting a card that is not a heart	39 (all remaining cards are non-hearts)	39	39/39 = 1 (Certain)
(iv)	Getting the Ace of Hearts	0 (hearts were all removed)	39	0/39 = 0 (Impossible)

Let's walk through the most important parts in detail:

Part (i) — Getting an Ace

Why is it 3/39 and not 4/39?

A full deck has 4 Aces — one in each suit. When all 13 heart cards are removed, the Ace of Hearts is also removed. Only 3 Aces remain: the Ace of Diamonds (♦), Ace of Spades (♠), and Ace of Clubs (♣).

n(S) = 39 (total remaining cards) n(A) = 3 (remaining aces: ♦A, ♠A, ♣A) P(A) = n(A)/n(S) = 3/39 = 1/13

Part (iii) — Not a Heart (Certain Event)

Why is the probability 1?

Since all 13 heart cards have been removed, every single one of the 39 remaining cards is not a heart. There is no card in the remaining pack that could be a heart. Picking any card guarantees it is not a heart — this is a certain event.

n(C) = 39 (every card remaining is non-heart) P(C) = 39/39 = 1

Question 7 — Rolling a Die

When a fair, six-sided die is rolled once, the sample space is S = {1, 2, 3, 4, 5, 6}, so n(S) = 6.

Highlighted faces show the prime numbers: 2, 3, 5

Part (i) — Getting a Prime Number

Prime numbers on a die: 2, 3, 5

A prime number has exactly two factors: 1 and itself. From 1–6, the primes are 2, 3, and 5. (Note: 1 is not a prime number.)

A = {2, 3, 5} → n(A) = 3 P(A) = 3/6 = 1/2

Part (ii) — A Number Between 2 and 6

"Between 2 and 6" means strictly between — not including 2 or 6

The numbers strictly between 2 and 6 are 3, 4, and 5. The endpoints 2 and 6 themselves are excluded.

B = {3, 4, 5} → n(B) = 3 P(B) = 3/6 = 1/2

Part (iii) — Getting an Odd Number

Odd numbers on a die: 1, 3, 5

C = {1, 3, 5} → n(C) = 3 P(C) = 3/6 = 1/2

📌 Interesting result: All three parts of Q7 give the same answer — 1/2. This is not a coincidence; exactly half of the six die faces satisfy each of these three different conditions.

Question 8 — Red King from a Full Deck of Cards

Problem 8

What is the probability of selecting a red king from a full deck of 52 cards?

A standard deck has two red suits: Hearts (♥) and Diamonds (♦). Each suit has one King. So there are exactly 2 red Kings in the deck: the King of Hearts and the King of Diamonds.

♥ King of Hearts ♦ King of Diamonds

n(S) = 52 (total cards in deck) n(A) = 2 (King of ♥ and King of ♦) P(A) = n(A)/n(S) = 2/52 = 1/26

All Answers — Quick Summary Table

Q#	Problem	Answer
1(i)	P(E) + P(not E)	1
1(ii)	Event that cannot happen	P = 0 — Impossible event
1(iii)	Event that is certain	P = 1 — Sure/Certain event
1(iv)	Sum of all elementary event probabilities	1
1(v)	Range of P(E)	0 ≤ P(E) ≤ 1
2(i)	Car starts / doesn't start	No — not equally likely
2(ii)	Basketball shot made / missed	No — not equally likely
2(iii)	True-false answer right / wrong	Yes — equally likely
2(iv)	Baby born boy / girl	Yes — equally likely
3	P(E) = 0.05 → P(not E)	0.95
4(i)	P(orange candy from lemon-only bag)	0 — Impossible
4(ii)	P(lemon candy from lemon-only bag)	1 — Certain
5(i)	P(Ace) from 39 cards	3/39 = 1/13
5(ii)	P(Diamond) from 39 cards	13/39 = 1/3
5(iii)	P(not a heart) from 39 cards	1 — Certain
5(iv)	P(Ace of Hearts) from 39 cards	0 — Impossible
6	P(same birthday), given P(not same) = 0.992	0.008
7(i)	P(prime) on a die	3/6 = 1/2
7(ii)	P(between 2 and 6) on a die	3/6 = 1/2
7(iii)	P(odd number) on a die	3/6 = 1/2
8	P(red king) from full deck	2/52 = 1/26

Common Mistakes to Avoid

"Between" does not include the endpoints: In Q7(ii), numbers between 2 and 6 are 3, 4, 5 — not 2, 3, 4, 5, 6. The word "between" is strictly exclusive unless "inclusive" or "from...to" is stated.
Forgetting that 1 is not a prime number: In Q7(i), students often include 1 and write A = {1, 2, 3, 5}, giving a wrong answer of 4/6. Prime numbers from 1–6 are only 2, 3, and 5.
Not adjusting the deck size in Q5: After hearts are removed, the sample space is 39, not 52. Using n(S) = 52 for all parts is the most common error in card probability questions.
Confusing "not a heart" with "a spade": "Not a heart" includes all three remaining suits — diamonds, spades, and clubs combined (39 cards). It is not just one suit.
Saying equally likely just because there are two outcomes: Q2 parts (i) and (ii) show that having exactly two outcomes does not automatically make them equally likely. The conditions that generate those outcomes must also be equal.
Wrong complementary event calculation: Always subtract from 1, not from 100. P(Ē) = 1 − P(E), not 100 − P(E).

⛔ Exam alert (CBSE, Telangana & AP boards): Playing card questions (like Q5 and Q8) appear almost every year. Know the structure of a standard deck — 52 cards, 4 suits of 13 each, 2 red suits (♥ ♦), 2 black suits (♠ ♣), 4 Aces, 4 Kings, 4 Queens, 4 Jacks — before attempting these questions.

What This Exercise Prepares You For

Exercise 13.1 builds the foundational language of probability — sample space, events, favourable outcomes, equally likely outcomes, and complementary events. These ideas are used in every subsequent probability problem, including multi-step and conditional scenarios.

The concepts of sample space and events connect back to set theory from the Statistics chapter. For more probability practice, the next exercises cover coins, two dice, and combined events. The complementary event rule from Q3 and Q6 is also heavily used in Exercise 13.2.

📐 Board Exam Tip (CBSE, Telangana & AP): Probability questions in the board exam are typically worth 2–4 marks. Exercise 13.1 problems — especially card and die questions — form the backbone of these marks. Master the three-step method: (1) write n(S), (2) list the favourable outcomes and find n(E), (3) calculate P(E) = n(E)/n(S) and simplify the fraction.