CBSE Class 10 Maths — Probability

Chapter Overview & Weightage

Probability is one of the most predictable chapters in CBSE Class 10 Maths — pun intended. Questions are consistently 4–6 marks per year, and the difficulty rarely exceeds “straightforward application of the basic formula.” A well-prepared student should aim to score full marks here.

Key Concepts You Must Know

Experiment: A process that produces a definite outcome. Example: tossing a coin, rolling a die.

Random experiment: An experiment where the outcome cannot be predicted with certainty in advance.

Sample space (S): The set of all possible outcomes of a random experiment.

• Tossing a coin: S = {H, T}, n(S) = 2
• Rolling a die: S = {1, 2, 3, 4, 5, 6}, n(S) = 6
• Tossing two coins: S = {HH, HT, TH, TT}, n(S) = 4

Event (E): A subset of the sample space — one or more outcomes of interest.

Probability of an event:

Key properties:

• $0 \leq P(E) \leq 1$ always
• $P(\text{impossible event}) = 0$
• $P(\text{certain event}) = 1$
• $P(E) + P(\bar{E}) = 1$, where $\bar{E}$ (or $E'$) is the complementary event

Important Formulas

Solved Previous Year Questions

PYQ 1 — Cards (CBSE 2022 type)

Q: One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is (i) a king (ii) a red card (iii) neither an ace nor a king.

Solution:

Total outcomes = 52

(i) Kings = 4. $P(\text{king}) = \frac{4}{52} = \frac{1}{13}$

(ii) Red cards = 26. $P(\text{red}) = \frac{26}{52} = \frac{1}{2}$

(iii) Aces = 4, Kings = 4. Cards that ARE aces or kings = 8. Cards that are NEITHER = 52 − 8 = 44. $P = \frac{44}{52} = \frac{11}{13}$

PYQ 2 — Dice (CBSE 2023 type)

Q: Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the two dice is (i) 7 (ii) a prime number.

Solution:

Total outcomes when two dice are thrown = $6 \times 6 = 36$.

(i) Sum = 7: Pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 pairs. $P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$

(ii) Sum is prime: Possible sums range from 2 to 12. Prime sums: 2, 3, 5, 7, 11.

• Sum 2: (1,1) → 1 pair
• Sum 3: (1,2), (2,1) → 2 pairs
• Sum 5: (1,4), (2,3), (3,2), (4,1) → 4 pairs
• Sum 7: 6 pairs (from above)
• Sum 11: (5,6), (6,5) → 2 pairs Total favourable = 1 + 2 + 4 + 6 + 2 = 15 pairs. $P = \frac{15}{36} = \frac{5}{12}$

PYQ 3 — Balls in a Bag (CBSE 2024 type)

Q: A bag contains 3 red, 5 white, and 4 black balls. One ball is drawn at random. What is the probability that the ball drawn is (i) red (ii) not black?

Solution:

Total balls = 3 + 5 + 4 = 12.

(i) $P(\text{red}) = \frac{3}{12} = \frac{1}{4}$

(ii) Not black = red + white = 3 + 5 = 8 balls. $P(\text{not black}) = \frac{8}{12} = \frac{2}{3}$

(Alternative: $P(\text{not black}) = 1 - P(\text{black}) = 1 - \frac{4}{12} = 1 - \frac{1}{3} = \frac{2}{3}$)

Difficulty Distribution

Expert Strategy

Step 1 — Draw the sample space if needed. For two coins or two dice, list all outcomes systematically. A $6 \times 6$ grid for two dice helps you count without missing outcomes.

Step 2 — Always simplify fractions. $\frac{4}{52}$ should be written as $\frac{1}{13}$. Leaving unsimplified fractions loses 0.5 marks.

Step 3 — Use complementary probability as a shortcut. If the question asks “probability of NOT getting a red card,” it’s faster to compute $1 - P(\text{red}) = 1 - 1/2 = 1/2$ rather than counting all 26 black cards.

Step 4 — Show the formula. Write $P(E) = \frac{n(E)}{n(S)}$ explicitly before substituting. Process marks are awarded for showing the setup.

Common Traps