NCERT Solutions for Class 10 Maths Chapter 14: Probability — Free PDF

Chapter 14 introduces the classical (theoretical) definition of probability and applies it to problems involving dice, coins, playing cards, and real-life situations. This is a highly scoring chapter because the problems follow predictable patterns and the arithmetic is straightforward.

The chapter has 2 exercises covering:
- Exercise 14.1: Classical probability with equally likely outcomes — coins, dice, bags of balls, and number-based problems
- Exercise 14.2: Mixed problems with playing cards, two-dice scenarios, multiple coins, and real-world word problems

This chapter typically carries 4-6 marks in the CBSE board exam. Playing card problems are the most frequently tested, followed by two-dice problems. The key to scoring full marks is systematic listing of outcomes and careful counting. Since every probability problem ultimately reduces to a simple fraction, careless errors in counting are the biggest risk.

Probability connects to the Statistics chapter (Chapter 13) and also lays the foundation for probability theory in Class 11 and 12.

Probability of an event:

Key properties:
- $0 \le P(E) \le 1$ for any event $E$
- $P(\text{sure event}) = 1$ (event that will definitely happen)
- $P(\text{impossible event}) = 0$ (event that cannot happen)
- $P(E) + P(\bar{E}) = 1$ (complementary events)
- $P(\bar{E}) = 1 - P(E)$ — this is the complement rule, very useful when counting "not" events is easier than counting favourable events

Common sample spaces:
- One coin: $\{H, T\}$ — $2$ outcomes
- Two coins: $\{HH, HT, TH, TT\}$ — $4$ outcomes
- Three coins: $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ — $8$ outcomes
- One die: $\{1, 2, 3, 4, 5, 6\}$ — $6$ outcomes
- Two dice: All ordered pairs $(a, b)$ where $a, b \in \{1,2,3,4,5,6\}$ — $36$ outcomes

**Standard deck of $52$ playing cards:**
- $4$ suits: Hearts, Diamonds (red), Spades, Clubs (black)
- $13$ cards per suit: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K
- $26$ red cards, $26$ black cards
- $4$ Aces, $4$ Kings, $4$ Queens, $4$ Jacks
- Face cards (J, Q, K): $12$ total (Aces are NOT face cards)
- Number cards (2 through 10): $36$ total

Two-dice sum table (for reference):

**Problem 1: A bag contains $3$ red and $5$ black balls. A ball is drawn at random. Find $P(\text{red})$ and $P(\text{not red})$.**

Solution:
Total balls $= 3 + 5 = 8$.

$P(\text{red}) = \dfrac{3}{8}$

$P(\text{not red}) = 1 - P(\text{red}) = 1 - \dfrac{3}{8} = \dfrac{5}{8}$

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**Problem 2: A die is thrown once. Find $P(\text{prime number})$ and $P(\text{number between 2 and 6})$.**

Solution:
Sample space $= \{1, 2, 3, 4, 5, 6\}$, total outcomes $= 6$.

(i) Prime numbers in the sample space $= \{2, 3, 5\}$, favourable outcomes $= 3$.

(ii) Numbers between $2$ and $6$ (exclusive) $= \{3, 4, 5\}$, favourable outcomes $= 3$.

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**Problem 3: A box contains $5$ red marbles, $8$ white, and $4$ green. One marble is drawn at random. Find the probability of each colour.**

Solution:
Total marbles $= 5 + 8 + 4 = 17$.

Verification: $\dfrac{5}{17} + \dfrac{8}{17} + \dfrac{4}{17} = \dfrac{17}{17} = 1$ ✓

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**Problem 4: A bag contains $12$ balls of which $x$ are white. If $6$ more white balls are added, the probability of drawing a white ball doubles. Find $x$.**

Solution:
Original probability of white $= \dfrac{x}{12}$.

After adding $6$ white: total $= 18$, white $= x + 6$.
New probability $= \dfrac{x+6}{18}$.

Given: new probability $= 2 \times$ original probability:

Answer: $x = 3$.

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**Problem 5: A jar contains $24$ marbles — some green and the rest blue. If a marble is drawn at random, $P(\text{green}) = \dfrac{2}{3}$. How many blue marbles are there?**

Solution:
Number of green marbles $= \dfrac{2}{3} \times 24 = 16$.

Number of blue marbles $= 24 - 16 = 8$.

Answer: $8$ blue marbles.

**Problem 1: A card is drawn from a well-shuffled deck of $52$ cards. Find $P(\text{king})$, $P(\text{not a king})$, $P(\text{face card})$.**

Solution:
Total cards $= 52$.

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

(ii) $P(\text{not king}) = 1 - \dfrac{1}{13} = \dfrac{12}{13}$

(iii) Face cards (J, Q, K) $= 12$: $P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13}$

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**Problem 2: Find $P(\text{red queen})$, $P(\text{black card})$, $P(\text{not a jack})$.**

Solution:

(i) Red queens $= 2$ (Queen of Hearts + Queen of Diamonds):

(ii) Black cards (Spades + Clubs) $= 26$:

(iii) Jacks $= 4$:

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Problem 3: Find the probability of getting a card that is neither a king nor a queen.

Solution:
Kings $= 4$, Queens $= 4$, Total kings or queens $= 8$.

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**Problem 4: Find $P(\text{spade or ace})$.**

Solution:
Spades $= 13$, Aces $= 4$. But the Ace of Spades is counted in both groups.

Favourable $= 13 + 4 - 1 = 16$.

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**Problem 5: Five cards — the ten, jack, queen, king, and ace of diamonds — are well-shuffled. One card is drawn. Find $P(\text{queen})$ and $P(\text{not an ace})$.**

Solution:
Total cards $= 5$.

$P(\text{queen}) = \dfrac{1}{5}$

$P(\text{not ace}) = 1 - \dfrac{1}{5} = \dfrac{4}{5}$

**Problem 1: Two dice are thrown simultaneously. Find $P(\text{sum} = 8)$.**

Solution:
Total outcomes $= 6 \times 6 = 36$.

Favourable outcomes for sum $= 8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$ — that is $5$ outcomes.

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**Problem 2: Two dice are thrown. Find $P(\text{doublet})$ and $P(\text{sum} > 10)$.**

Solution:

(i) Doublets: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ — $6$ outcomes.

(ii) Sum $> 10$ means sum $= 11$ or $12$.
Sum $= 11$: $(5,6), (6,5)$ — $2$ outcomes.
Sum $= 12$: $(6,6)$ — $1$ outcome.

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**Problem 3: Three coins are tossed simultaneously. Find $P(\text{at least 2 heads})$.**

Solution:
Sample space: $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ — $8$ outcomes.

At least $2$ heads: $\{HHH, HHT, HTH, THH\}$ — $4$ outcomes.

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**Problem 4: Two dice are thrown. Find $P(\text{product of numbers is } 12)$.**

Solution:
Pairs where the product is $12$: $(2,6), (3,4), (4,3), (6,2)$ — $4$ outcomes.

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**Problem 5: A coin is tossed $3$ times. Find $P(\text{exactly one head})$.**

Solution:
Total outcomes $= 8$.

Exactly one head: $\{HTT, THT, TTH\}$ — $3$ outcomes.

**Example 1: A number is selected at random from the first $20$ natural numbers. Find $P(\text{divisible by 3 or 5})$.**

Solution:
Numbers from $1$ to $20$.

Divisible by $3$: $\{3, 6, 9, 12, 15, 18\}$ — $6$ numbers.
Divisible by $5$: $\{5, 10, 15, 20\}$ — $4$ numbers.
Divisible by both $3$ and $5$ (i.e., by $15$): $\{15\}$ — $1$ number.

By inclusion-exclusion: $6 + 4 - 1 = 9$.

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**Example 2: A game consists of tossing a coin $3$ times and noting the outcomes. Hanif wins if all tosses give the same result. What is his probability of losing?**

Solution:
Total outcomes $= 8$.

All same: $\{HHH, TTT\}$ — $2$ outcomes. $P(\text{win}) = \dfrac{2}{8} = \dfrac{1}{4}$.

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**Example 3: Two dice are thrown. Find $P(\text{sum is a perfect square})$.**

Solution:
Possible sums range from $2$ to $12$. Perfect squares in this range: $4$ and $9$.

Sum $= 4$: $(1,3), (2,2), (3,1)$ — $3$ outcomes.
Sum $= 9$: $(3,6), (4,5), (5,4), (6,3)$ — $4$ outcomes.

Total favourable $= 3 + 4 = 7$.

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**Example 4: A box contains $90$ discs numbered $1$ to $90$. One disc is drawn at random. Find $P(\text{two-digit number})$.**

Solution:
Two-digit numbers from $1$ to $90$: $10, 11, 12, \ldots, 90$ — that is $81$ numbers.

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**Example 5: Cards numbered $1$ to $20$ are put in a box. A card is drawn at random. Find $P(\text{prime number})$.**

Solution:
Primes from $1$ to $20$: $\{2, 3, 5, 7, 11, 13, 17, 19\}$ — $8$ primes.

**Mistake 1: Forgetting that $(3, 5)$ and $(5, 3)$ are different outcomes for two dice.**
When two dice are thrown, the order matters. There are $36$ outcomes, not $21$. The outcome $(3, 5)$ means die 1 shows $3$ and die 2 shows $5$, while $(5, 3)$ is the reverse. Both must be counted separately.

Mistake 2: Including Aces as face cards.
Aces are NOT face cards. Face cards are only Jacks, Queens, and Kings — $12$ total. If a question asks for the probability of drawing a face card, use $\dfrac{12}{52}$, not $\dfrac{16}{52}$.

Mistake 3: Not simplifying the final fraction.
Always reduce fractions to lowest terms. Write $\dfrac{1}{13}$ instead of $\dfrac{4}{52}$. CBSE examiners may deduct marks for unsimplified answers.

**Mistake 4: Getting a probability greater than $1$.**
If your answer is greater than $1$, something is wrong. Probability is always between $0$ and $1$ (inclusive). Go back and check your counting.

Mistake 5: Double-counting in "or" problems.
When asked for $P(A \text{ or } B)$, if $A$ and $B$ can overlap, you must subtract the overlap: favourable $= |A| + |B| - |A \cap B|$. For example, $P(\text{spade or ace}) = \dfrac{13 + 4 - 1}{52}$ because the Ace of Spades is both a spade and an ace.

Q1. A die is thrown once. Find $P(\text{even number})$ and $P(\text{number} \geq 3)$.

Answer: Even numbers: $\{2,4,6\}$, so $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$. Numbers $\geq 3$: $\{3,4,5,6\}$, so $P(\geq 3) = \dfrac{4}{6} = \dfrac{2}{3}$.

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Q2. A card is drawn from a deck of $52$ cards. Find $P(\text{red face card})$.

Answer: Red face cards: J, Q, K of hearts and J, Q, K of diamonds $= 6$. $P = \dfrac{6}{52} = \dfrac{3}{26}$.

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Q3. Two dice are thrown. Find $P(\text{sum} = 7)$.

Answer: Favourable: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6$. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

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Q4. A bag contains $4$ red, $5$ blue, and $3$ green balls. Find $P(\text{not blue})$.

Answer: Total $= 12$. Not blue $= 4 + 3 = 7$. $P(\text{not blue}) = \dfrac{7}{12}$.

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Q5. Numbers $1$ to $15$ are written on slips and put in a box. One slip is drawn. Find $P(\text{multiple of 4})$.

Answer: Multiples of $4$: $\{4, 8, 12\} = 3$ numbers. $P = \dfrac{3}{15} = \dfrac{1}{5}$.

• Probability of an event $= \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$, and always lies between $0$ and $1$.
  - The complement rule $P(\bar{E}) = 1 - P(E)$ is the most useful shortcut in probability.
  - For two dice, there are always $36$ outcomes. The most likely sum is $7$ (with $6$ ways).
  - A standard deck has $52$ cards, $12$ face cards (Aces are NOT face cards), $26$ red and $26$ black.
  - For three coins, there are $8$ outcomes. For $n$ coins, there are $2^n$ outcomes.
  - Always list outcomes systematically to avoid miscounting.
  - This is one of the most scoring chapters — with practice, full marks are very achievable.

Practise probability problems on SparkEd for exam-style questions!