Class 10 Maths Chapter 14: Probability Notes & Formulas

Exam Weightage & Blueprint

Total: ~4-5 Marks

This chapter falls under Unit VII: Statistics & Probability (11 marks total). As per the latest syllabus: Classical definition of probability. Simple problems on finding the probability of an event.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	Very High	Basic Probability Formula, Dice/Cards
Short Answer	2 or 3	High	2 Dice problems, Card properties
Case Study	4	Medium	Real-life scenarios (Balls in bag, Spinning wheel)

CBSE 2026–27 Syllabus Status: Probability

Unit VII: Statistics & Probability • 4–5 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

📐 Important Definitions & Formulas

1. Theoretical Probability

The theoretical probability of an event E is defined as:

$$ P(E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes} $$

(Pierre Simon Laplace, 1795)

2. Complementary Events

The event representing "not E" is called the complement of E, denoted by $\overline{E}$.

$$ P(E) + P(\overline{E}) = 1 \quad \Rightarrow \quad P(\overline{E}) = 1 - P(E) $$

3. Range of Probability

Rule: The probability of an event lies between 0 and 1 (inclusive).
$$ 0 \le P(E) \le 1 $$

Sure Event: Probability is 1 (Certain to happen).
Impossible Event: Probability is 0 (Cannot happen).

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Equally Likely Outcomes Condition: The classical formula $P(E) = \frac{m}{n}$ is valid only if all elementary outcomes are equally likely. (e.g. "a driver attempts to start a car" is not equally likely).
Sum of Elementary Events: The sum of the probabilities of all elementary events of an experiment is strictly $1$: $$\sum P(E_i) = P(E_1) + P(E_2) + \dots + P(E_n) = 1$$
Terminology Meanings:
• "At least $k$": $\ge k$ (value $k$ or more).
• "At most $k$": $\le k$ (value $k$ or less).
• "Neither A nor B": $1 - P(A \cup B)$.
Calendar Probabilities (53 Sundays / Mondays):
• Ordinary year (365 days = 52 weeks + 1 day) $\implies P(53\text{ Sundays}) = \mathbf{\frac{1}{7}}$.
• Leap year (366 days = 52 weeks + 2 days) $\implies P(53\text{ Sundays}) = \mathbf{\frac{2}{7}}$.
Cards Removed Subset: If $k$ cards are removed from the deck before drawing, total outcomes becomes $(52 - k)$ instead of 52.

🎯 Mastering Sample Spaces

Most errors happen because students count total outcomes incorrectly. Memorize these standard experiments.

1. 💰 Coins

Experiment	Total Outcomes	Sample Space
1 Coin	2	{H, T}
2 Coins	4	{HH, HT, TH, TT}
3 Coins	8	{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

[Image of coin probability tree diagram]

2. 🎲 Dice

1 Die: Outcomes are {1, 2, 3, 4, 5, 6}. Total = 6.
Prime Numbers: 2, 3, 5 (Note: 1 is NOT prime).

2 Dice: Total Outcomes = $6 \times 6 = 36$.
Important for sums (e.g., Sum=8: (2,6), (3,5), (4,4), (5,3), (6,2)).

3. 🃏 Playing Cards

Total Cards = 52. Divided into 4 Suits of 13 cards each.

Color	Suit	Cards
Red (26)	Hearts ?, Diamonds ?	A, 2-10, J, Q, K
Black (26)	Spades ?, Clubs ?	A, 2-10, J, Q, K

Face Cards: Kings, Queens, Jacks. Total = 12 (3 per suit). Aces are NOT face cards!

Solved Examples (Board Marking Scheme)

Q1. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will be (i) be an ace, (ii) not be an ace. (2 Marks)

Step 1: Identify Total Outcomes 0.5 Mark

Total number of cards = 52.

Step 2: Probability of Ace (E) 0.5 Mark

Number of Aces = 4.

$P(E) = \frac{4}{52} = \frac{1}{13}$.

Step 3: Probability of Not Ace (F) 1 Mark

$P(F) = 1 - P(E) = 1 - \frac{1}{13} = \frac{12}{13}$.

Alternatively: Non-ace cards = $52 - 4 = 48$. $P(F) = $\frac{48}{52}$ = $\frac{12}{13}$$.

Q2. Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is 8? (2 Marks)

Step 1: Total Outcomes 0.5 Mark

Total outcomes when throwing 2 dice = $6 \times 6 = 36$.

Step 2: Favourable Outcomes 1 Mark

Pairs with sum 8: $(2,6), (3,5), (4,4), (5,3), (6,2)$.

Number of favourable outcomes = 5.

Step 3: Calculation 0.5 Mark

$P(sum is 8) = \frac{5}{36}$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Probability Range: Never write an answer greater than 1 (e.g., 1.5) or negative. It is an instant 0 marks.

Prime Numbers: In dice problems, students often count 1 as a prime number. Remember: Primes are 2, 3, 5.

"At least": "At least one Head" in 2 coins means {HT, TH, HH} (3 outcomes), not just 1.

💡 Scoring Tips

Simplify Fractions: Always simplify your final answer (e.g., write $\frac{1}{2}$ instead of $\frac{26}{52}$).

Show Sample Space: For 2-3 mark questions, explicitly write the favourable outcomes (e.g., "Favourable outcomes: {2, 3, 5}").

Sum Rule: The sum of probabilities of all elementary events of an experiment is 1. Use this to check your answers.

📝 More Solved Board Questions

Q3. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red color, (ii) a red face card. 3 Marks

Sol. Total Cards = 52.

(i) Kings of Red color = 2 (King of Hearts and King of Diamonds).

$P(Red King) = 2/52 = \mathbf{1/26}$.

(ii) Face cards per suit = 3. Suits of red color = 2.

Total red face cards = $3 \times 2 = 6$.

$P(Red Face Card) = 6/52 = \mathbf{3/26}$.

🎯 Board Pattern (2018–2025): Questions often specify color AND type (e.g., "Black Queen" or "Red Ace"). Always count carefully: 26 red, 26 black; 13 per suit; 4 per rank.

📋 Board Revision Checklist

✅ $P(E) = Fav outcomes / Total outcomes$
✅ Range: $0 \le P(E) \le 1$
✅ $P(E) + P(not E) = 1$
✅ Sure event = 1; Impossible event = 0
✅ 2 Coins: 4 outcomes; 3 Coins: 8 outcomes
✅ 2 Dice: 36 outcomes
✅ Ordinary year: $P(53\text{ Sundays}) = 1/7$; Leap year: $P(53\text{ Sundays}) = 2/7$
✅ Sum of probabilities of all elementary events $= 1$
✅ Cards removed: total outcomes becomes $52 - k$ (do not divide by 52)
✅ 52 Cards: 12 face cards, 4 suits, 26 red/black
✅ "At least" means that value or MORE
✅ "At most" means that value or LESS

💡 Exam Tip:
If you are unsure of the sample space for 2 dice, quickly draw the $6 \times 6$ table on your rough sheet. It takes 1 minute but guarantees accuracy for sum-related questions.

CBSE Section E • 4 Marks

Case Study: Quality Control in Industrial Fasteners

Scenario: A precision machine component manufacturer packs steel locking pins into lot containers. A carton contains 90 identical circular steel washers numbered from 1 to 90. One washer is drawn at random from the carton during a random quality audit.

Q1: Find the total number of elementary outcomes in the sample space $n(S)$. 1 Mark

Q2: Find the probability that the drawn washer bears a two-digit number. 1 Mark

Q3: Find the probability that the drawn washer bears: (a) a perfect square number, and (b) a number divisible by 5. (OR) Find the probability that the number drawn is a prime number less than 20. 2 Marks

📝 10-Mark Self-Assessment Test

Time: 25 Mins

Solve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.

Q1 (1 Mark - MCQ): If $P(E) = 0.07$, then the probability of 'not $E$' is:
(a) $0.93$ (b) $0.95$ (c) $0.03$ (d) $0.07$

Q2 (2 Marks - VSA): Two coins are tossed simultaneously. Find the probability of getting:
(i) at least one head, (ii) at most one head.

Q3 (3 Marks - SA): A box contains 90 discs numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears:
(i) a two-digit number,
(ii) a perfect square number,
(iii) a number divisible by 5.

Q4 (4 Marks - LA): All kings, queens, and aces are removed from a well-shuffled pack of 52 playing cards. From the remaining cards, one card is drawn at random. Find the probability that the card drawn is:
(i) a black face card,
(ii) a red card,
(iii) a ten of spades,
(iv) a picture card.

Sol 1: $P(\bar{E}) = 1 - P(E) = 1 - 0.07 = 0.93$.
Correct Option: (a) $0.93$ [1 Mark]

Sol 2: Sample space $S = \{HH, HT, TH, TT\}$, Total $= 4$.
(i) At least one head: $\{HH, HT, TH\} \implies P = \frac{3}{4}$ [1 Mark].
(ii) At most one head (0 or 1 head): $\{HT, TH, TT\} \implies P = \frac{3}{4}$ [1 Mark].
Answers: (i) $3/4$, (ii) $3/4$ [2 Marks]

Sol 3: Total discs $= 90$.
(i) Two-digit numbers (10 to 90): $90 - 9 = 81$ numbers.
$$P = \frac{81}{90} = \frac{9}{10} [1 Mark]$$ (ii) Perfect squares ($1, 4, 9, 16, 25, 36, 49, 64, 81$): 9 numbers.
$$P = \frac{9}{90} = \frac{1}{10} [1 Mark]$$ (iii) Numbers divisible by 5 ($5, 10, 15, \dots, 90$): $90/5 = 18$ numbers.
$$P = \frac{18}{90} = \frac{1}{5} [1 Mark]$$ Answers: (i) $9/10$, (ii) $1/10$, (iii) $1/5$ [3 Marks]

Sol 4: Cards removed: 4 Kings + 4 Queens + 4 Aces $= 12$ cards.
Remaining cards in pack $= 52 - 12 = 40$ cards.
(i) Black face cards remaining: Only Jacks remain ($J\spadesuit, J\clubsuit$) $= 2$ cards.
$$P = \frac{2}{40} = \frac{1}{20} [1 Mark]$$ (ii) Red cards remaining: Originally 26 red cards. Removed: 2 red kings, 2 red queens, 2 red aces $= 6$ cards. Remaining red $= 26 - 6 = 20$.
$$P = \frac{20}{40} = \frac{1}{2} [1 Mark]$$ (iii) Ten of spades: Only 1 such card.
$$P = \frac{1}{40} [1 Mark]$$ (iv) Picture cards (face cards) remaining: Only 4 Jacks remain.
$$P = \frac{4}{40} = \frac{1}{10} [1 Mark]$$ Answers: (i) $1/20$, (ii) $1/2$, (iii) $1/40$, (iv) $1/10$ [4 Marks]

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. Which of the following cannot be the probability of an event?

2. An event is very unlikely to happen. Its probability is closest to:

3. If a card is selected from a deck of 52 cards, the probability of its being a red face card is:

4. A die is thrown once. The probability of getting a prime number is:

5. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, number of blue balls is:

Frequently Asked Questions: Class 10 Maths Chapter 14

What is the marks weightage of Probability in Class 10 Maths?

Probability carries 4 to 5 marks in the board exam, typically distributed as 1 or 2 MCQs and a 2-mark or 3-mark problem on cards, dice, or coins.

What is the theoretical probability formula and its range?

P(E) = (Number of outcomes favorable to E) / (Total number of possible outcomes). The probability of any event satisfies 0 <= P(E) <= 1. The sum of probabilities of all elementary events is 1.

What are complementary events in probability?

The event not-E is the complement of event E, and P(E) + P(not-E) = 1, or P(not-E) = 1 - P(E).

What is the standard 52-card deck distribution tested in CBSE exams?

A standard deck has 52 cards divided into 4 suits (13 Spades, 13 Clubs which are black; 13 Hearts, 13 Diamonds which are red). There are 12 face cards (4 Kings, 4 Queens, 4 Jacks) and 4 Aces.

Chapter 14 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 14: Probability

Overview