Is it correct to say: Treating (1, 2) and (2, 1) as the same outcome when two dice are thrown.?

The outcomes are ordered pairs; there are 36 of them. A sum of 3 can come in 2 ways: (1, 2) and (2, 1).

Is it correct to say: Taking 3 outcomes for two coins (two heads, two tails, one of each) and giving each probability 1/3.?

These three are not equally likely. The equally likely outcomes are HH, HT, TH, TT, so P(one of each) = 2/4 = 1/2.

Is it correct to say: Counting aces as face cards.?

Face cards are only J, Q, K: 3 per suit, 12 in all.

Is it correct to say: Counting 1 as a prime number.?

1 is neither prime nor composite. Primes from 1 to 6 are 2, 3, 5.

Is it correct to say: Using 52 as the total after some cards have been removed.?

Recount: if all red face cards are removed, 52 − 6 = 46 cards remain.

Is it correct to say: Getting a probability greater than 1, or adding probabilities of events that overlap.?

Check 0 ≤ P(E) ≤ 1. For 'heart or king', count the king of hearts only once.

📄 Formula sheet (PDF)

Probability

Theoretical (classical) probability of an event, complementary events, sure and impossible events, with problems on coins, dice, playing cards and random draws — NCERT Class 10 Maths Ch 14

📐 10 formulas✏️ 4 examples🎯 6 practice⚖️ Unit VII (Statistics and Probability): 11 marks (with Ch 13)🏫 CBSE📚 Class 10✓ 2025–26 syllabus

💡

Board Exam Tips

→Write the number of possible outcomes and the number of favourable outcomes separately, then form P(E). List the favourable outcomes when there are only a few.
→Give the answer as a fraction in lowest terms. Any probability must lie between 0 and 1.
→For two dice, think of the 36 outcomes as a 6 × 6 grid of ordered pairs; (1, 2) and (2, 1) are different outcomes.
→Know the pack: 52 cards, 4 suits of 13, 26 red and 26 black, 12 face cards (J, Q, K), 4 aces.
→For a 'not E' question, P(not E) = 1 − P(E) is usually quicker than counting.
→If some cards or balls are removed, recount the total before finding any probability.

📐 Formulas(10)

Theoretical Probability of an Event★ Board fav

P (E) = \frac{number of outcomes favourable to E}{number of all possible outcomes of the experiment}

Symbol	Meaning
$E$	The event whose probability is required
$P (E)$	Probability of E (a number from 0 to 1, no unit)

Range of Probability

0 \leq P (E) \leq 1

Sure Event and Impossible Event

P (sure event) = 1, P (impossible event) = 0

Complementary Events★ Board fav

P (E) + P (E) = 1 \Rightarrow P (not E) = 1 - P (E)

Symbol	Meaning
$E$	Complement of E, the event 'not E'

Sum over Elementary Events

P (E_{1}) + P (E_{2}) + \dots + P (E_{n}) = 1

Tossing Coins

outcomes = 2^{k} for k coins : 2, 4, 8

Throwing Dice

one die: 6 outcomes, two dice: 6 \times 6 = 36 outcomes

Sum of the Numbers on Two Dice

ways to get sum s = 6 - ∣ s - 7∣, s = 2, 3, \dots, 12

A Pack of 52 Playing Cards★ Board fav

52 = 4 suits \times 13, face cards = 12, red = black = 26

Counting Multiples from 1 to N

multiples of k in 1, 2, \dots, N = ⌊ \frac{N}{k} ⌋

✏️ Solved Examples

1Solved Exampleeasy3 steps

A die is thrown once. Find the probability of getting (i) a prime number (ii) a number greater than 4.

Possible outcomes: 1, 2, 3, 4, 5, 6, so there are 6 equally likely outcomes.

2Solved Exampleboard4 steps

Two dice are thrown at the same time. Find the probability that (i) the sum of the numbers is 8 (ii) the sum is not 8 (iii) the outcome is a doublet.

Total outcomes = 6 × 6 = 36 ordered pairs.

3Solved Exampleboard3 steps

One card is drawn from a well-shuffled pack of 52 playing cards. Find the probability that it is (i) a red face card (ii) neither a heart nor a king.

Total outcomes = 52.

4Solved ExampleHOTS4 steps

A jar contains only red, blue and green marbles. A marble is drawn at random. The probability of red is 1/4 and the probability of blue is 1/3. If the jar has 10 green marbles, find the total number of marbles and the number of red and blue marbles.

Every marble is exactly one of red, blue or green, so the probabilities of these three events add up to 1.

P (green) = 1 - \frac{1}{4} - \frac{1}{3} = \frac{12 - 3 - 4}{12} = \frac{5}{12}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Treating (1, 2) and (2, 1) as the same outcome when two dice are thrown.
✓The outcomes are ordered pairs; there are 36 of them. A sum of 3 can come in 2 ways: (1, 2) and (2, 1).
2
Taking 3 outcomes for two coins (two heads, two tails, one of each) and giving each probability 1/3.
✓These three are not equally likely. The equally likely outcomes are HH, HT, TH, TT, so P(one of each) = 2/4 = 1/2.
3
Counting aces as face cards.
✓Face cards are only J, Q, K: 3 per suit, 12 in all.
4
Counting 1 as a prime number.
✓1 is neither prime nor composite. Primes from 1 to 6 are 2, 3, 5.
5
Using 52 as the total after some cards have been removed.
✓Recount: if all red face cards are removed, 52 − 6 = 46 cards remain.
6
Getting a probability greater than 1, or adding probabilities of events that overlap.
✓Check 0 ≤ P(E) ≤ 1. For 'heart or king', count the king of hearts only once.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A bag contains 5 red, 8 white and 7 green balls. One ball is drawn at random. Find the probability that it is not green.
Q2
Three coins are tossed together. Find the probability of getting at least two heads.
Q3
Two dice are thrown. Find the probability that the product of the numbers is 12.
Q4
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is not a face card.
Q5
Discs numbered 1 to 50 are put in a box and one is drawn at random. Find the probability that its number is divisible by 3 or 5.
Q6
Find the probability that a leap year chosen at random has 53 Sundays.

📝 Notes

Probability

This chapter uses the theoretical (classical) definition of probability. If an experiment has equally likely outcomes, the probability of an event is the number of favourable outcomes divided by the total number of outcomes.

Three steps for every question

Count all outcomes. Write the sample space or its size: 6 for a die, 36 for two dice, 52 for a pack of cards.
Count the favourable outcomes. List them when there are only a few; this shows the examiner your reasoning.
Divide and simplify. The answer must lie between 0 and 1.

Complementary events

E and "not E" together cover every outcome, so $P(E) + P(\overline{E}) = 1$ . Use the complement when "not E" is quicker to count, or when the question gives P(E) and asks for P(not E). The same idea extends to all elementary events: their probabilities add to 1. Likewise, when every outcome falls in exactly one of a few events (red, blue or green in the marble example), their probabilities add to 1, which lets you find a missing probability.

Standard experiments to know

Coins: 1 coin gives 2 outcomes, 2 coins give 4, 3 coins give 8.
Dice: two dice give 36 ordered pairs. Sum 7 is the most likely total (6 ways).
Cards: 4 suits × 13 cards. Hearts and diamonds are red, spades and clubs are black. Each suit has 3 face cards (J, Q, K) and one ace.
Numbers 1 to 100: there are 25 primes; 1 is neither prime nor composite.

Sanity checks

P(sure event) = 1 and P(impossible event) = 0.
If a question says "at random" or "well-shuffled", the outcomes are equally likely and the classical formula applies.