Focus Keyword: NCERT Solutions Class 10 Math Chapter 14 Probability
Secondary Keywords & LSI: CBSE Class 10 Maths Probability solutions, Class 10 Maths Chapter 14 Exercise 14.1, theoretical probability Class 10, probability of playing cards two dice problems, CBSE Class 10 Maths board exam preparation
SEO Meta Description: Complete NCERT Solutions for Class 10 Math Chapter 14 Probability. Step-by-Step solutions for all Exercise 14.1 questions, card decks, dice, & CBSE tips.
H1 Title: NCERT Solutions for Class 10 Math Chapter 14: Probability (Complete Step-by-Step Guide)
Navigating through the CBSE Class 10 Mathematics curriculum requires an in-depth understanding of theoretical probability, equally likely outcomes, complementary events, and sample spaces involving coins, dice, and playing cards. Chapter 14 of Class 10 Mathematics, “Probability”, forms the foundation of risk analysis, statistical inference, data science algorithms, and decision-making theory. It investigates the classical definition of probability pioneered by Pierre-Simon Laplace; explores the deterministic boundaries of impossible and sure events; and details the quantitative determination of likelihoods across single-stage and multi-stage random experiments. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from foundational textbook examples to the complete 25-question set of Exercise 14.1 and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Probability), and clear geometric and sample space visualisations have been highlighted to ensure students secure maximum marks in their CBSE Board Examinations.
Chapter 14: Probability
Master Chapter Summary & Quantitative Blueprint
In the rationalised NCERT Class 10 curriculum, Chapter 14 focuses entirely on classical (theoretical) probability, assuming all elementary outcomes of an experiment are equally likely.
| Concept / Experiment | Mathematical Definition | Governing Formula / Identity | Key Boundary Condition | CBSE Marks Weightage |
|---|---|---|---|---|
| Classical Probability | Ratio of favourable outcomes to total outcomes | $P(E) = \frac{n(E)}{n(S)}$ | $0 \le P(E) \le 1$ | 1 to 2 Marks |
| Complementary Event | The event ‘not $E$’, denoted by $\bar{E}$ or $E’$ | $P(E) + P(\bar{E}) = 1$ | $P(\bar{E}) = 1 – P(E)$ | 1 to 2 Marks |
| Impossible Event | An event that cannot happen under any condition | $P(\phi) = 0$ | Favourable outcomes $= 0$ | 1 Mark (MCQ) |
| Sure / Certain Event | An event that is guaranteed to happen | $P(S) = 1$ | Favourable outcomes $= n(S)$ | 1 Mark (MCQ) |
| Playing Cards (Deck) | 52 cards: 26 Red (Hearts, Diamonds), 26 Black (Spades, Clubs) | 4 suits of 13 cards each; 12 Face Cards | Face cards: 4 Kings, 4 Queens, 4 Jacks | 2 to 3 Marks |
| Two Dice Experiment | Rolling two balanced dice simultaneously | Total outcomes: $6 \times 6 = 36$ | Outcomes: $(1,1)$ to $(6,6)$ | 3 to 4 Marks |
| Coin Toss (1, 2, 3) | Fair coin flipped $n$ times | Total outcomes: $2^n$ ($2, 4, 8$) | For 3 coins: $n(S) = 8$ | 2 to 3 Marks |
🧠 Examiner’s Secret: A probability value can never be negative, nor can it ever exceed $1$. If your final answer simplifies to a number less than $0$, greater than $1$, or a percentage exceeding $100%$, an arithmetic inversion has occurred. Always write probabilities as simplest fractions (e.g., $\frac{1}{2}$ instead of $\frac{2}{4}$) or decimals.
Foundational Probability Concepts and Sample Spaces
Theoretical Probability
Theoretical (classical) probability of an event $E$ is the ratio of the number of outcomes favourable to $E$ to the total number of all possible outcomes of the experiment.
$$P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes of the experiment}} = \frac{n(E)}{n(S)}$$
This definition relies on the fundamental assumption that all outcomes of the experiment are equally likely (each individual outcome has the exact same chance of occurring).
Complementary Events and Probability Range
For any event $E$:
- The probability of an impossible event is $0$.
- The probability of a sure or certain event is $1$.
- The probability of any arbitrary event satisfies the double inequality:
$$0 \le P(E) \le 1$$ - The event representing the non-occurrence of $E$ is called its complementary event, denoted $\bar{E}$:
$$P(E) + P(\bar{E}) = 1 \implies P(\bar{E}) = 1 – P(E)$$
Standard 52-Card Deck Breakdown
A standard deck contains $52$ playing cards divided into $4$ suits of $13$ cards each: Total Playing Cards: 52 26 Red Cards 13 Hearts (♥) + 13 Diamonds (♦) 26 Black Cards 13 Spades (♠) + 13 Clubs (♣) 12 Face Cards (Picture Cards) 4 Kings, 4 Queens, 4 Jacks (6 Red, 6 Black) | Aces are NOT face cards
Two-Dice Experiment Sample Space Grid ($n(S) = 36$)
When two dice (e.g., one blue and one grey) are rolled simultaneously, the complete sample space consists of $6 \times 6 = 36$ equally likely ordered pairs:
$$\begin{matrix}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)
\end{matrix}$$
💡 Did You Know?: The minimum possible sum of two dice is $2$ (with $1$ favourable outcome: $(1,1)$), the maximum sum is $12$ (with $1$ favourable outcome: $(6,6)$), and the most probable sum is $7$ (with $6$ favourable outcomes: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$), yielding $P(\text{Sum } 7) = \frac{6}{36} = \frac{1}{6}$.
[👉 Also Read: Class 10 Math Chapter 13 Statistics NCERT Solutions]
Step-by-Step Solutions: NCERT Class 10 Mathematics Chapter 14 Solved Examples
Example 1 (Page 203) [CBSE 2018, 2021]
Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
Answer:
Step 1: Identify Sample Space
When a balanced coin is tossed once, the possible outcomes are Head ($H$) and Tail ($T$).
$$S = {H, T} \implies n(S) = 2$$
Step 2: Probability of Getting a Head
Let $E$ be the event of getting a head: $E = {H} \implies n(E) = 1$.
$$P(E) = \frac{n(E)}{n(S)} = \frac{1}{2}$$
Step 3: Probability of Getting a Tail
Let $F$ be the event of getting a tail: $F = {T} \implies n(F) = 1$.
$$P(F) = \frac{n(F)}{n(S)} = \frac{1}{2}$$
Final Answer:
The probability of getting a head is $\frac{1}{2}$, and the probability of getting a tail is $\frac{1}{2}$.
Example 2 (Page 203) [CBSE 2017, 2020]
A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the:
(i) yellow ball?
(ii) red ball?
(iii) blue ball?
Answer:
Step 1: Identify Total Outcomes
There are $3$ balls of identical shape and size: $n(S) = 1 + 1 + 1 = 3$.
Step 2: Calculate Individual Probabilities
- (i) Let $Y$ be the event of drawing a yellow ball. Favourable outcome $n(Y) = 1$.
$$P(Y) = \frac{n(Y)}{n(S)} = \frac{1}{3}$$ - (ii) Let $R$ be the event of drawing a red ball. Favourable outcome $n(R) = 1$.
$$P(R) = \frac{n(R)}{n(S)} = \frac{1}{3}$$ - (iii) Let $B$ be the event of drawing a blue ball. Favourable outcome $n(B) = 1$.
$$P(B) = \frac{n(B)}{n(S)} = \frac{1}{3}$$
Notice that $P(Y) + P(R) + P(B) = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$.
Final Answer:
(i) $P(\text{yellow}) = \mathbf{\frac{1}{3}}$, (ii) $P(\text{red}) = \mathbf{\frac{1}{3}}$, (iii) $P(\text{blue}) = \mathbf{\frac{1}{3}}$.
Example 3 (Page 204) [CBSE 2016, 2019]
Suppose we throw a die once. (i) What is the probability of getting a number greater than $4$? (ii) What is the probability of getting a number less than or equal to $4$?
Answer:
Step 1: Identify Sample Space
$$S = {1, 2, 3, 4, 5, 6} \implies n(S) = 6$$
Step 2: (i) Probability of Getting a Number Greater Than 4
Let $E$ be the event of getting a number greater than $4$: $E = {5, 6} \implies n(E) = 2$.
$$P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$$
Step 3: (ii) Probability of Getting a Number Less Than or Equal to 4
Let $F$ be the event of getting a number $\le 4$: $F = {1, 2, 3, 4} \implies n(F) = 4$.
$$P(F) = \frac{n(F)}{n(S)} = \frac{4}{6} = \frac{2}{3}$$
Alternatively, since $F = \bar{E}$:
$$P(F) = 1 – P(E) = 1 – \frac{1}{3} = \frac{2}{3}$$
Final Answer:
(i) $P(\text{number } > 4) = \mathbf{\frac{1}{3}}$, (ii) $P(\text{number } \le 4) = \mathbf{\frac{2}{3}}$.
Example 4 (Page 205) [CBSE 2015, 2020 Standard]
One card is drawn from a well-shuffled deck of $52\text{ cards}$. Calculate the probability that the card will:
(i) be an ace,
(ii) not be an ace.
Answer:
Step 1: Identify Total Outcomes
Total number of cards in a standard deck, $n(S) = 52$.
Step 2: (i) Probability of Drawing an Ace
There are $4$ aces in a deck (one in each suit: Spade, Club, Heart, Diamond): $n(E) = 4$.
$$P(\text{Ace}) = \frac{n(E)}{n(S)} = \frac{4}{52} = \frac{1}{13}$$
Step 3: (ii) Probability of Not Drawing an Ace
Using the complementary event formula:
$$P(\text{Not Ace}) = 1 – P(\text{Ace}) = 1 – \frac{1}{13} = \frac{12}{13}$$
Alternatively, the number of non-ace cards is $52 – 4 = 48$:
$$P(\text{Not Ace}) = \frac{48}{52} = \frac{12}{13}$$
Final Answer:
(i) $P(\text{Ace}) = \mathbf{\frac{1}{13}}$, (ii) $P(\text{Not Ace}) = \mathbf{\frac{12}{13}}$.
Example 5 (Page 205) [CBSE 2014, 2018, 2023]
Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is $0.62$. What is the probability of Reshma winning the match?
Answer:
Step 1: Characterise Complementary Outcomes
In a two-player tennis match, if Sangeeta wins, Reshma loses, and vice versa. There are no ties.
Let $S$ be the event that Sangeeta wins, and $R$ be the event that Reshma wins.
The events $S$ and $R$ are complementary: $R = \bar{S}$.
Step 2: Compute Probability
$$P(S) = 0.62$$
$$P(R) = 1 – P(S) = 1 – 0.62 = 0.38$$
Final Answer:
The probability of Reshma winning the match is $0.38$.
Example 6 (Page 206) [CBSE 2016, 2020 Standard]
Savita and Hamida are friends. What is the probability that both will have:
(i) different birthdays?
(ii) the same birthday? (ignoring a leap year).
Answer:
Step 1: Identify Sample Space
In a non-leap year, there are $365$ days.
The total number of possible pairs of birthdays for two individuals is $365 \times 365$.
Step 2: (i) Probability of Having Different Birthdays
Hamida’s birthday can be any day of the year. For Savita to have a different birthday, her birthday can fall on any of the remaining $364$ days:
$$P(\text{different birthdays}) = \frac{364}{365}$$
Step 3: (ii) Probability of Having the Same Birthday
Having the same birthday is the complement of having different birthdays:
$$P(\text{same birthday}) = 1 – P(\text{different birthdays}) = 1 – \frac{364}{365} = \frac{1}{365}$$
Final Answer:
(i) $P(\text{different birthdays}) = \mathbf{\frac{364}{365}}$, (ii) $P(\text{same birthday}) = \mathbf{\frac{1}{365}}$.
Example 7 (Page 207) [CBSE 2017, 2022 Term-2]
There are $40\text{ students}$ in Class X of a school of whom $25$ are girls and $15$ are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of:
(i) a girl?
(ii) a boy?
Answer:
Step 1: Identify Total Outcomes
Total number of students, $n(S) = 25 + 15 = 40$.
Step 2: Calculate Probabilities
- (i) Let $G$ be the event of selecting a girl’s name. Favourable outcomes $n(G) = 25$.
$$P(G) = \frac{n(G)}{n(S)} = \frac{25}{40} = \frac{5}{8}$$ - (ii) Let $B$ be the event of selecting a boy’s name. Favourable outcomes $n(B) = 15$.
$$P(B) = \frac{n(B)}{n(S)} = \frac{15}{40} = \frac{3}{8}$$
Alternatively: $P(B) = 1 – P(G) = 1 – \frac{5}{8} = \frac{3}{8}$.
Final Answer:
(i) $P(\text{Girl}) = \mathbf{\frac{5}{8}}$, (ii) $P(\text{Boy}) = \mathbf{\frac{3}{8}}$.
Example 8 (Page 207) [CBSE 2013, 2019 Set-1]
A box contains $3$ blue, $2$ white, and $4$ red marbles. If a marble is drawn at random from the box, what is the probability that it will be:
(i) white?
(ii) blue?
(iii) red?
Answer:
Step 1: Identify Total Outcomes
Total number of marbles, $n(S) = 3 + 2 + 4 = 9$.
Step 2: Calculate Probabilities
- (i) Let $W$ be the event of drawing a white marble: $n(W) = 2$.
$$P(W) = \frac{n(W)}{n(S)} = \frac{2}{9}$$ - (ii) Let $B$ be the event of drawing a blue marble: $n(B) = 3$.
$$P(B) = \frac{n(B)}{n(S)} = \frac{3}{9} = \frac{1}{3}$$ - (iii) Let $R$ be the event of drawing a red marble: $n(R) = 4$.
$$P(R) = \frac{n(R)}{n(S)} = \frac{4}{9}$$
Final Answer:
(i) $P(\text{white}) = \mathbf{\frac{2}{9}}$, (ii) $P(\text{blue}) = \mathbf{\frac{1}{3}}$, (iii) $P(\text{red}) = \mathbf{\frac{4}{9}}$.
Example 9 (Page 208) [CBSE 2015, 2020 Standard]
Harpreet tosses two different coins simultaneously (say, one is of ₹$1$ and other of ₹$2$). What is the probability that she gets at least one head?
Answer:
Step 1: Identify Sample Space
$$S = {(H,H), (H,T), (T,H), (T,T)} \implies n(S) = 4$$
Step 2: Define Favourable Event
“At least one head” means $1$ head or $2$ heads.
Favourable outcomes: $E = {(H,H), (H,T), (T,H)} \implies n(E) = 3$.
Step 3: Compute Probability
$$P(E) = \frac{n(E)}{n(S)} = \frac{3}{4}$$
Alternatively, using the complementary event (no heads = $(T,T)$):
$$P(\text{at least one head}) = 1 – P(\text{no heads}) = 1 – \frac{1}{4} = \frac{3}{4}$$
Final Answer:
The probability of getting at least one head is $\frac{3}{4}$.
Example 10 (Page 209) [CBSE 2014, 2018]
In a musical chair game, the person playing the music has been advised to stop playing the music at any time within $2\text{ minutes}$ after she starts playing. What is the probability that the music will stop within the first half-minute?
Answer:
Step 1: Geometric / Continuous Probability Formulation
Total possible time interval for music to stop = $[0, 2]\text{ minutes}$.
Total duration = $2\text{ minutes}$.
Step 2: Favourable Time Interval
The music stops within the first half-minute ($0.5\text{ minutes}$ or $\frac{1}{2}\text{ minute}$).
Favourable interval length = $0.5\text{ minutes}$.
Step 3: Compute Probability
$$P(E) = \frac{\text{Length of favourable interval}}{\text{Total length of interval}} = \frac{0.5}{2} = \frac{1}{4}$$
Final Answer:
The probability that the music stops within the first half-minute is $\frac{1}{4}$ (or $0.25$).
Example 11 (Page 209) [CBSE 2016, 2019 Set-2]
A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig. 14.2. What is the probability that it crashed inside the lake shown in the figure? (The rectangular region is $9\text{ km} \times 4.5\text{ km}$; the lake is located in a corner of dimensions $2.5\text{ km} \times 2\text{ km}$).
Answer:
Step 1: Calculate Total Area of Rectangular Region
$$\text{Total Area} = \text{Length} \times \text{Breadth} = 9\text{ km} \times 4.5\text{ km} = 40.5\text{ km}^2$$
Step 2: Calculate Area of the Lake
$$\text{Area of Lake} = \text{Length} \times \text{Breadth} = (9 – 6)\text{ km} \text{ or given as } 2.5\text{ km} \times 2\text{ km} = 5\text{ km}^2$$
(From NCERT Fig. 14.2: lake width $= 9 – 6.5 = 2.5\text{ km}$, lake height $= 4.5 – 2.5 = 2\text{ km}$. Area $= 2.5 \times 2 = 5\text{ km}^2$)
Step 3: Compute Geometric Probability
$$P(\text{Crashed in lake}) = \frac{\text{Area of the lake}}{\text{Total area of the rectangular region}} = \frac{5}{40.5} = \frac{50}{405} = \frac{10}{81}$$
Final Answer:
The probability that the helicopter crashed inside the lake is $\frac{10}{81}$.
Example 12 (Page 210) [CBSE 2015, 2023 Set-1]
A carton consists of $100\text{ shirts}$ of which $88$ are good, $8$ have minor defects and $4$ have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that:
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?
Answer:
Step 1: Identify Total Outcomes
Total number of shirts, $n(S) = 100$.
Step 2: (i) Acceptable to Jimmy
Jimmy accepts only good shirts.
Number of good shirts, $n(J) = 88$.
$$P(\text{Acceptable to Jimmy}) = \frac{88}{100} = 0.88 = \frac{22}{25}$$
Step 3: (ii) Acceptable to Sujatha
Sujatha rejects only shirts with major defects, meaning she accepts good shirts and shirts with minor defects.
Number of acceptable shirts, $n(\text{Sujatha}) = 88 + 8 = 96$ (or $100 – 4 = 96$).
$$P(\text{Acceptable to Sujatha}) = \frac{96}{100} = 0.96 = \frac{24}{25}$$
Final Answer:
(i) $P(\text{Jimmy}) = \mathbf{0.88}$ (or $\mathbf{\frac{22}{25}}$), (ii) $P(\text{Sujatha}) = \mathbf{0.96}$ (or $\mathbf{\frac{24}{25}}$).
Example 13 (Page 211) [CBSE 2014, 2018, 2020 Standard, 2024]
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is:
(i) $8$?
(ii) $13$?
(iii) less than or equal to $12$?
Answer:
Step 1: Identify Total Outcomes
Total possible outcomes = $6 \times 6 = 36$.
Step 2: (i) Probability that Sum is 8
Favourable outcomes where sum $= 8$:
$$E = {(2,6), (3,5), (4,4), (5,3), (6,2)} \implies n(E) = 5$$
$$P(\text{Sum } 8) = \frac{n(E)}{n(S)} = \frac{5}{36}$$
Step 3: (ii) Probability that Sum is 13
The maximum possible sum on two dice is $6 + 6 = 12$. No outcome can yield a sum of $13$.
Favourable outcomes $= 0$ (Impossible event).
$$P(\text{Sum } 13) = \frac{0}{36} = 0$$
Step 4: (iii) Probability that Sum is Less Than or Equal to 12
Every pair of numbers on two standard dice satisfies $1 + 1 \le \text{Sum} \le 6 + 6$, meaning all sums fall between $2$ and $12$.
All $36$ outcomes are favourable (Sure event).
$$P(\text{Sum } \le 12) = \frac{36}{36} = 1$$
Final Answer:
(i) $P(\text{Sum } 8) = \mathbf{\frac{5}{36}}$, (ii) $P(\text{Sum } 13) = \mathbf{0}$, (iii) $P(\text{Sum } \le 12) = \mathbf{1}$.
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 14.1
Question 1 (Page 212) [CBSE 2012, 2017 Objective]
Complete the following statements:
(i) Probability of an event $E$ + Probability of the event ‘not $E$’ = _______.
(ii) The probability of an event that cannot happen is _______. Such an event is called _______.
(iii) The probability of an event that is certain to happen is _______. Such an event is called _______.
(iv) The sum of the probabilities of all the elementary events of an experiment is _______.
(v) The probability of an event is greater than or equal to _______ and less than or equal to _______.
Answer:
(i) $1$ [Complementary event rule: $P(E) + P(\bar{E}) = 1$]
(ii) $0$ ; impossible event
(iii) $1$ ; sure event (or certain event)
(iv) $1$ [$\sum P(E_i) = 1$]
(v) $0$ ; $1$ [$0 \le P(E) \le 1$]
Question 2 (Page 212) [CBSE 2015, 2019 Concept]
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Answer:
(i) Not equally likely. Whether a car starts depends on its mechanical condition, battery, and fuel. Starting and not starting do not share an equal chance under normal conditions.
(ii) Not equally likely. A player’s success depends on athletic skill, training, defense, and distance; shooting or missing are not inherently equally probable.
(iii) Equally likely. There are only two options (True or False), and guessing gives an equal $\frac{1}{2}$ chance of being correct or incorrect.
(iv) Equally likely. Biologically, assuming standard random birth outcomes, a newborn is equally likely to be a boy or a girl ($P = \frac{1}{2}$).
Question 3 (Page 212) [CBSE 2014, 2018]
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Answer:
Tossing a fair coin is considered completely unbiased and fair because:
- The experiment has only two mutually exclusive outcomes: Head or Tail.
- The outcomes are equally likely ($P(H) = P(T) = \frac{1}{2}$).
- The result of a coin toss is unpredictable and cannot be manipulated by either team.
Question 4 (Page 212) [CBSE 2016, 2020 Standard Objective]
Which of the following cannot be the probability of an event?
(A) $\frac{2}{3}$
(B) $-1.5$
(C) $15%$
(D) $0.7$
Answer:
The correct option is (B) $-1.5$.
Reason:
By the fundamental axioms of probability, the probability of any event $E$ must satisfy $0 \le P(E) \le 1$. A probability value can never be negative.
- (A) $\frac{2}{3} \approx 0.67 \in [0,1]$
- (C) $15% = 0.15 \in [0,1]$
- (D) $0.7 \in [0,1]$
- (B) $-1.5 < 0$, which is impossible.
Question 5 (Page 212) [CBSE 2013, 2017, 2023 Set-1]
If $P(E) = 0.05$, what is the probability of ‘not $E$’?
Answer:
Step 1: Apply Complementary Event Identity
$$P(\text{not } E) = P(\bar{E}) = 1 – P(E)$$
Step 2: Substitution
$$P(\text{not } E) = 1 – 0.05 = 0.95$$
Final Answer:
The probability of ‘not $E$’ is $0.95$.
Question 6 (Page 212) [CBSE 2015, 2018]
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Answer:
Step 1: (i) Orange Flavoured Candy
The bag contains only lemon flavoured candies. There are no orange candies present ($n(E) = 0$).
$$P(\text{Orange candy}) = \frac{0}{\text{Total candies}} = 0$$
This is an impossible event.
Step 2: (ii) Lemon Flavoured Candy
Every candy in the bag is lemon flavoured. Whichever candy is picked will always be lemon flavoured ($n(E) = n(S)$).
$$P(\text{Lemon candy}) = \frac{n(S)}{n(S)} = 1$$
This is a sure (certain) event.
Final Answer:
(i) $P(\text{orange}) = \mathbf{0}$, (ii) $P(\text{lemon}) = \mathbf{1}$.
Question 7 (Page 212) [CBSE 2014, 2019 Set-1]
It is given that in a group of $3\text{ students}$, the probability of $2\text{ students}$ not having the same birthday is $0.992$. What is the probability that the $2\text{ students}$ have the same birthday?
Answer:
Step 1: Identify Complementary Relationship
Let $E$ be the event that 2 students have the same birthday.
Then $\bar{E}$ is the event that 2 students do not have the same birthday.
Given: $P(\bar{E}) = 0.992$.
Step 2: Calculate $P(E)$
$$P(E) = 1 – P(\bar{E}) = 1 – 0.992 = 0.008$$
Final Answer:
The probability that the two students have the same birthday is $0.008$.
Question 8 (Page 212) [CBSE 2012, 2016, 2022 Term-2]
A bag contains $3\text{ red balls}$ and $5\text{ black balls}$. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) red?
(ii) not red?
Answer:
Step 1: Identify Total Outcomes
Total number of balls, $n(S) = 3 + 5 = 8$.
Step 2: (i) Probability of Red Ball
Number of red balls, $n(R) = 3$.
$$P(\text{Red}) = \frac{n(R)}{n(S)} = \frac{3}{8}$$
Step 3: (ii) Probability of Not Red Ball
Using complementary events:
$$P(\text{Not Red}) = 1 – P(\text{Red}) = 1 – \frac{3}{8} = \frac{5}{8}$$
(Alternatively, a ball that is not red must be black: $\frac{5}{8}$).
Final Answer:
(i) $P(\text{red}) = \mathbf{\frac{3}{8}}$, (ii) $P(\text{not red}) = \mathbf{\frac{5}{8}}$.
Question 9 (Page 213) [CBSE 2015, 2018, 2023 Set-2]
A box contains $5\text{ red marbles}$, $8\text{ white marbles}$ and $4\text{ green marbles}$. One marble is taken out of the box at random. What is the probability that the marble taken out will be:
(i) red?
(ii) white?
(iii) not green?
Answer:
Step 1: Identify Total Outcomes
Total marbles, $n(S) = 5 + 8 + 4 = 17$.
Step 2: Calculate Probabilities
- (i) Red marbles, $n(R) = 5$:
$$P(\text{Red}) = \frac{5}{17}$$ - (ii) White marbles, $n(W) = 8$:
$$P(\text{White}) = \frac{8}{17}$$ - (iii) Not green means red or white: $n(\text{Not Green}) = 5 + 8 = 13$:
$$P(\text{Not Green}) = \frac{13}{17}$$
(Alternatively: $1 – P(\text{Green}) = 1 – \frac{4}{17} = \frac{13}{17}$).
Final Answer:
(i) $P(\text{red}) = \mathbf{\frac{5}{17}}$, (ii) $P(\text{white}) = \mathbf{\frac{8}{17}}$, (iii) $P(\text{not green}) = \mathbf{\frac{13}{17}}$.
Question 10 (Page 213) [CBSE 2014, 2017, 2020 Standard]
A piggy bank contains hundred $50\text{p}$ coins, fifty ₹$1$ coins, twenty ₹$2$ coins and ten ₹$5$ coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin:
(i) will be a $50\text{p}$ coin?
(ii) will not be a ₹$5$ coin?
Answer:
Step 1: Identify Total Number of Coins
- $50\text{p}$ coins = $100$
- ₹$1$ coins = $50$
- ₹$2$ coins = $20$
- ₹$5$ coins = $10$
- Total coins, $n(S) = 100 + 50 + 20 + 10 = 180$.
Step 2: (i) Probability of a 50p Coin
Favourable coins, $n(50\text{p}) = 100$.
$$P(50\text{p}) = \frac{100}{180} = \frac{10}{18} = \frac{5}{9}$$
Step 3: (ii) Probability that Coin will Not be a ₹5 Coin
Number of ₹$5$ coins = $10$.
Number of coins that are not ₹$5$ = $180 – 10 = 170$.
$$P(\text{Not ₹5}) = \frac{170}{180} = \frac{17}{18}$$
(Alternatively: $1 – P(₹5) = 1 – \frac{10}{180} = 1 – \frac{1}{18} = \frac{17}{18}$).
Final Answer:
(i) $P(50\text{p}) = \mathbf{\frac{5}{9}}$, (ii) $P(\text{not ₹5}) = \mathbf{\frac{17}{18}}$.
Question 11 (Page 213) [CBSE 2013, 2018]
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing $5\text{ male fish}$ and $8\text{ female fish}$ (see Fig. 14.4). What is the probability that the fish taken out is a male fish?
Answer:
Step 1: Identify Total Outcomes
- Male fish = $5$
- Female fish = $8$
- Total fish in tank, $n(S) = 5 + 8 = 13$.
Step 2: Probability of Male Fish
Favourable outcomes, $n(\text{Male}) = 5$.
$$P(\text{Male Fish}) = \frac{n(\text{Male})}{n(S)} = \frac{5}{13}$$
Final Answer:
The probability that the fish taken out is a male fish is $\frac{5}{13}$.
Question 12 (Page 213) [CBSE 2016, 2020 Standard, 2024]
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1, 2, 3, 4, 5, 6, 7, 8$ (see Fig. 14.5), and these are equally likely outcomes. What is the probability that it will point at:
(i) $8$?
(ii) an odd number?
(iii) a number greater than $2$?
(iv) a number less than $9$?
Answer:
Step 1: Identify Sample Space
$$S = {1, 2, 3, 4, 5, 6, 7, 8} \implies n(S) = 8$$
Step 2: Calculate Probabilities
- (i) Pointing at $8$: Favourable outcome = ${8} \implies n(E) = 1$.
$$P(8) = \frac{1}{8}$$ - (ii) An odd number: Favourable outcomes = ${1, 3, 5, 7} \implies n(E) = 4$.
$$P(\text{Odd}) = \frac{4}{8} = \frac{1}{2}$$ - (iii) A number greater than $2$: Favourable outcomes = ${3, 4, 5, 6, 7, 8} \implies n(E) = 6$.
$$P(> 2) = \frac{6}{8} = \frac{3}{4}$$ - (iv) A number less than $9$: All numbers in the sample space are less than $9$: $n(E) = 8$.
$$P(< 9) = \frac{8}{8} = 1 \quad (\text{Sure Event})$$
Final Answer:
(i) $\mathbf{\frac{1}{8}}$, (ii) $\mathbf{\frac{1}{2}}$, (iii) $\mathbf{\frac{3}{4}}$, (iv) $\mathbf{1}$.
Question 13 (Page 213) [CBSE 2015, 2019 Set-3]
A die is thrown once. Find the probability of getting:
(i) a prime number;
(ii) a number lying between $2$ and $6$;
(iii) an odd number.
Answer:
Step 1: Identify Sample Space
$$S = {1, 2, 3, 4, 5, 6} \implies n(S) = 6$$
Step 2: Calculate Probabilities
- (i) Prime numbers on a die: ${2, 3, 5}$ (Note: 1 is neither prime nor composite).
Favourable outcomes $n(E) = 3$.
$$P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$$ - (ii) Numbers strictly between $2$ and $6$: ${3, 4, 5}$.
Favourable outcomes $n(E) = 3$.
$$P(\text{Between 2 and 6}) = \frac{3}{6} = \frac{1}{2}$$ - (iii) Odd numbers: ${1, 3, 5}$.
Favourable outcomes $n(E) = 3$.
$$P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$$
Final Answer:
(i) $P(\text{prime}) = \mathbf{\frac{1}{2}}$, (ii) $P(\text{between 2 and 6}) = \mathbf{\frac{1}{2}}$, (iii) $P(\text{odd}) = \mathbf{\frac{1}{2}}$.
Question 14 (Page 213) [CBSE 2012, 2017, 2020 Standard, 2023]
One card is drawn from a well-shuffled deck of $52\text{ cards}$. Find the probability of getting:
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Answer:
Step 1: Identify Total Outcomes
Total cards, $n(S) = 52$.
Step 2: Calculate Each Probability
- (i) King of red colour: King of Hearts, King of Diamonds $\implies n = 2$.
$$P(\text{Red King}) = \frac{2}{52} = \frac{1}{26}$$ - (ii) A face card: Kings, Queens, Jacks ($4 \times 3 = 12$ cards) $\implies n = 12$.
$$P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}$$ - (iii) A red face card: 3 Hearts face cards + 3 Diamonds face cards $\implies n = 6$.
$$P(\text{Red Face Card}) = \frac{6}{52} = \frac{3}{26}$$ - (iv) The jack of hearts: Exactly $1$ card in the deck $\implies n = 1$.
$$P(\text{Jack of Hearts}) = \frac{1}{52}$$ - (v) A spade: $13$ cards in the spade suit $\implies n = 13$.
$$P(\text{Spade}) = \frac{13}{52} = \frac{1}{4}$$ - (vi) The queen of diamonds: Exactly $1$ card in the deck $\implies n = 1$.
$$P(\text{Queen of Diamonds}) = \frac{1}{52}$$
Final Answer:
(i) $\mathbf{\frac{1}{26}}$, (ii) $\mathbf{\frac{3}{13}}$, (iii) $\mathbf{\frac{3}{26}}$, (iv) $\mathbf{\frac{1}{52}}$, (v) $\mathbf{\frac{1}{4}}$, (vi) $\mathbf{\frac{1}{52}}$.
Question 15 (Page 214) [CBSE 2014, 2018, 2022 Term-2]
Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Answer:
Step 1: (i) Probability of Queen initially
Initial set of cards = ${10, J, Q, K, A}$ of diamonds $\implies n(S_1) = 5$.
Number of queens = $1$.
$$P(\text{Queen}) = \frac{1}{5}$$
Step 2: (ii) After Queen is Put Aside
The queen is removed and not replaced.
Remaining cards = ${10, J, K, A} \implies$ New sample space $n(S_2) = 4$.
- (a) Probability of an ace: There is $1$ ace among the $4$ cards:
$$P(\text{Ace}) = \frac{1}{4}$$ - (b) Probability of a queen: There are $0$ queens remaining:
$$P(\text{Queen}) = \frac{0}{4} = 0 \quad (\text{Impossible Event})$$
Final Answer:
(i) $\mathbf{\frac{1}{5}}$, (ii)(a) $\mathbf{\frac{1}{4}}$, (ii)(b) $\mathbf{0}$.
Question 16 (Page 214) [CBSE 2013, 2019 Set-1]
$12\text{ defective pens}$ are accidentally mixed with $132\text{ good ones}$. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer:
Step 1: Identify Total Outcomes
- Good pens = $132$
- Defective pens = $12$
- Total pens, $n(S) = 132 + 12 = 144$.
Step 2: Probability of a Good Pen
Favourable outcomes, $n(\text{Good}) = 132$.
$$P(\text{Good Pen}) = \frac{132}{144}$$
Divide numerator and denominator by $12$:
$$P(\text{Good Pen}) = \frac{132 \div 12}{144 \div 12} = \frac{11}{12}$$
Final Answer:
The probability that the pen taken out is a good one is $\frac{11}{12}$.
Question 17 (Page 214) [CBSE 2016, 2020 Standard]
(i) A lot of $20\text{ bulbs}$ contain $4\text{ defective ones}$. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer:
Part (i):
- Total bulbs, $n(S) = 20$.
- Defective bulbs = $4$.
$$P(\text{Defective}) = \frac{4}{20} = \frac{1}{5}$$
Part (ii):
- A non-defective bulb was drawn in part (i) and not replaced.
- Remaining non-defective bulbs = $(20 – 4) – 1 = 16 – 1 = 15$.
- Defective bulbs remaining = $4$.
- New total bulbs, $n(S’) = 20 – 1 = 19$.
$$P(\text{Not Defective}) = \frac{15}{19}$$
Final Answer:
(i) $P(\text{defective}) = \mathbf{\frac{1}{5}}$, (ii) $P(\text{not defective}) = \mathbf{\frac{15}{19}}$.
Question 18 (Page 214) [CBSE 2015, 2018, 2023 Set-3]
A box contains $90\text{ discs}$ which are numbered from $1$ 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 $5$.
Answer:
Step 1: Identify Total Outcomes
Numbers from $1$ to $90 \implies n(S) = 90$.
Step 2: (i) Probability of a Two-Digit Number
Single-digit numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9$ (total $9$).
Two-digit numbers = $90 – 9 = 81$ (from $10$ to $90$).
$$P(\text{Two-Digit}) = \frac{81}{90} = \frac{9}{10} = 0.9$$
Step 3: (ii) Probability of a Perfect Square Number
Perfect squares between $1$ and $90$:
$$1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81$$
Total perfect squares = $9$.
$$P(\text{Perfect Square}) = \frac{9}{90} = \frac{1}{10} = 0.1$$
Step 4: (iii) Probability of a Number Divisible by 5
Numbers divisible by $5$: $5, 10, 15, 20, \dots, 90$.
Number of terms: $\frac{90}{5} = 18$.
$$P(\text{Divisible by 5}) = \frac{18}{90} = \frac{1}{5} = 0.2$$
Final Answer:
(i) $\mathbf{\frac{9}{10}}$ (or $0.9$), (ii) $\mathbf{\frac{1}{10}}$ (or $0.1$), (iii) $\mathbf{\frac{1}{5}}$ (or $0.2$).
Question 19 (Page 214) [CBSE 2014, 2019 Set-2]
A child has a die whose six faces show the letters as given below:
| A | B | C | D | E | A |
The die is thrown once. What is the probability of getting:
(i) A?
(ii) D?
Answer:
Step 1: Identify Sample Space
Total faces, $n(S) = 6$.
Step 2: Calculate Probabilities
- (i) Face ‘A’ appears on $2$ faces: $n(A) = 2$.
$$P(A) = \frac{2}{6} = \frac{1}{3}$$ - (ii) Face ‘D’ appears on $1$ face: $n(D) = 1$.
$$P(D) = \frac{1}{6}$$
Final Answer:
(i) $P(A) = \mathbf{\frac{1}{3}}$, (ii) $P(D) = \mathbf{\frac{1}{6}}$.
Question 20 (Page 214) [CBSE 2013, 2017] (Non-Examinable in Board Exams per NCERT Footnote, provided for conceptual completeness)
Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter $1\text{ m}$? (The rectangular region has length $3\text{ m}$ and breadth $2\text{ m}$).
Answer:
Step 1: Total Area of Rectangular Target
$$\text{Area of Rectangle} = l \times b = 3\text{ m} \times 2\text{ m} = 6\text{ m}^2$$
Step 2: Area of Circular Region
Diameter $d = 1\text{ m} \implies$ Radius $r = \frac{1}{2}\text{ m} = 0.5\text{ m}$.
$$\text{Area of Circle} = \pi r^2 = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}\text{ m}^2$$
Step 3: Compute Geometric Probability
$$P(\text{Lands in Circle}) = \frac{\text{Area of Circle}}{\text{Area of Rectangle}} = \frac{\frac{\pi}{4}}{6} = \frac{\pi}{24}$$
Using $\pi \approx 3.14$: $\frac{3.14}{24} \approx 0.131$.
Final Answer:
The probability is $\frac{\pi}{24}$.
Question 21 (Page 215) [CBSE 2015, 2020 Standard]
A lot consists of $144\text{ ball pens}$ of which $20$ are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:
(i) She will buy it?
(ii) She will not buy it?
Answer:
Step 1: Identify Total Outcomes
- Total pens, $n(S) = 144$.
- Defective pens = $20$.
- Good pens = $144 – 20 = 124$.
Step 2: (i) Probability She Will Buy It
Nuri buys the pen only if it is good:
$$P(\text{Buys}) = \frac{124}{144} = \frac{31}{36}$$
Step 3: (ii) Probability She Will Not Buy It
Nuri does not buy if it is defective:
$$P(\text{Will not buy}) = \frac{20}{144} = \frac{5}{36}$$
(Alternatively: $1 – \frac{31}{36} = \frac{5}{36}$).
Final Answer:
(i) $P(\text{buys}) = \mathbf{\frac{31}{36}}$, (ii) $P(\text{will not buy}) = \mathbf{\frac{5}{36}}$.
Question 22 (Page 215) [CBSE 2012, 2016, 2024]
Refer to Example 13.
(i) Complete the following table:
| Event: ‘Sum on 2 dice’ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | $\frac{1}{36}$ | $\frac{5}{36}$ | $\frac{1}{36}$ |
(ii) A student argues that ‘there are 11 possible outcomes $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ and $12$. Therefore, each of them has a probability $\frac{1}{11}$’. Do you agree with this argument? Justify your answer.
Answer:
Part (i): Completed Table
Count the favourable outcomes for each sum from the $36$-outcome sample space:
- Sum = 2: $(1,1) \implies 1$ outcome $\implies \frac{1}{36}$
- Sum = 3: $(1,2), (2,1) \implies 2$ outcomes $\implies \frac{2}{36}$
- Sum = 4: $(1,3), (2,2), (3,1) \implies 3$ outcomes $\implies \frac{3}{36}$
- Sum = 5: $(1,4), (2,3), (3,2), (4,1) \implies 4$ outcomes $\implies \frac{4}{36}$
- Sum = 6: $(1,5), (2,4), (3,3), (4,2), (5,1) \implies 5$ outcomes $\implies \frac{5}{36}$
- Sum = 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \implies 6$ outcomes $\implies \frac{6}{36}$
- Sum = 8: $(2,6), (3,5), (4,4), (5,3), (6,2) \implies 5$ outcomes $\implies \frac{5}{36}$
- Sum = 9: $(3,6), (4,5), (5,4), (6,3) \implies 4$ outcomes $\implies \frac{4}{36}$
- Sum = 10: $(4,6), (5,5), (6,4) \implies 3$ outcomes $\implies \frac{3}{36}$
- Sum = 11: $(5,6), (6,5) \implies 2$ outcomes $\implies \frac{2}{36}$
- Sum = 12: $(6,6) \implies 1$ outcome $\implies \frac{1}{36}$
| Event: ‘Sum on 2 dice’ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | $\frac{1}{36}$ | $\frac{2}{36}$ | $\frac{3}{36}$ | $\frac{4}{36}$ | $\frac{5}{36}$ | $\frac{6}{36}$ | $\frac{5}{36}$ | $\frac{4}{36}$ | $\frac{3}{36}$ | $\frac{2}{36}$ | $\frac{1}{36}$ |
Part (ii): Evaluation of Argument
No, I do not agree with the argument.
The $11$ sums are not equally likely outcomes. For example, the sum $7$ can occur in $6$ different ways, whereas the sum $2$ can occur in only $1$ way. The classical probability formula applies only when the individual elementary outcomes are equally likely.
Question 23 (Page 215) [CBSE 2014, 2018, 2023 Set-1]
A game consists of tossing a one rupee coin $3\text{ times}$ and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Answer:
Step 1: Identify Sample Space for 3 Coins
$$S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} \implies n(S) = 2^3 = 8$$
Step 2: Identify Winning Outcomes
Hanif wins if he gets three heads or three tails:
$$W = {HHH, TTT} \implies n(W) = 2$$
$$P(\text{Hanif wins}) = \frac{2}{8} = \frac{1}{4}$$
Step 3: Calculate Probability That Hanif Loses
Losing is the complement of winning:
$$P(\text{Hanif loses}) = 1 – P(\text{Hanif wins}) = 1 – \frac{1}{4} = \frac{3}{4}$$
(Alternatively, there are $8 – 2 = 6$ losing outcomes: $\frac{6}{8} = \frac{3}{4}$).
Final Answer:
The probability that Hanif will lose the game is $\frac{3}{4}$.
Question 24 (Page 215) [CBSE 2013, 2017, 2020 Standard]
A die is thrown twice. What is the probability that:
(i) $5$ will not come up either time?
(ii) $5$ will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Answer:
Step 1: Total Sample Space
Total outcomes, $n(S) = 6 \times 6 = 36$.
Step 2: Identify Outcomes Where 5 Comes Up At Least Once
Outcomes having $5$ on the first die: $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$ ($6$ outcomes).
Outcomes having $5$ on the second die: $(1,5), (2,5), (3,5), (4,5), (6,5)$ ($5$ outcomes, avoiding double-counting $(5,5)$).
Total outcomes where $5$ comes up at least once:
$$n(E) = 6 + 5 = 11$$
Step 3: (i) Probability That 5 Will Not Come Up Either Time
Outcomes where $5$ does not appear on either die = $36 – 11 = 25$.
$$P(\text{5 will not come up}) = \frac{25}{36}$$
Step 4: (ii) Probability That 5 Will Come Up At Least Once
$$P(\text{5 comes up at least once}) = \frac{11}{36}$$
(Alternatively: $1 – \frac{25}{36} = \frac{11}{36}$).
Final Answer:
(i) $P(\text{no 5}) = \mathbf{\frac{25}{36}}$, (ii) $P(\text{at least one 5}) = \mathbf{\frac{11}{36}}$.
Question 25 (Page 215) [CBSE 2015, 2019 Concept]
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $\frac{1}{3}$.
(ii) If a die is thrown, there are two possible outcomes — an odd number or an even number. Therefore, the probability of getting an odd number is $\frac{1}{2}$.
Answer:
(i) Incorrect.
The three stated outcomes are not equally likely.
When two fair coins are tossed, the actual sample space has $4$ equally likely elementary outcomes:
$$S = {(H,H), (H,T), (T,H), (T,T)}$$
- $P(\text{two heads}) = P(H,H) = \frac{1}{4}$
- $P(\text{two tails}) = P(T,T) = \frac{1}{4}$
- $P(\text{one of each}) = P(H,T) + P(T,H) = \frac{2}{4} = \frac{1}{2}$
Because $P(\text{one of each}) = \frac{1}{2} \neq \frac{1}{4}$, the outcomes do not each have probability $\frac{1}{3}$.
(ii) Correct.
When a standard fair die is thrown, the sample space is ${1, 2, 3, 4, 5, 6}$.
- Odd numbers = ${1, 3, 5} \implies 3$ outcomes
- Even numbers = ${2, 4, 6} \implies 3$ outcomes
Both categories have the same number of equally likely outcomes ($3$ each).
Therefore, $P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$, making the argument mathematically valid.
Master High-Yield Board FAQs (Rank Math Schema Ready)
What is theoretical probability and how does it differ from experimental probability?
Theoretical probability evaluates the likelihood of an event based on mathematical reasoning and the assumption of equally likely outcomes ($P(E) = \frac{n(E)}{n(S)}$), without actually performing any physical trials. Experimental (empirical) probability is calculated from the actual observed results of a repeated experiment ($\frac{\text{Number of trials where event occurred}}{\text{Total number of trials performed}}$).
Can the probability of an event be negative or greater than 1?
No. The probability of any event $E$ is bounded within the closed interval $0 \le P(E) \le 1$. A probability of $0$ corresponds to an impossible event that cannot occur, while a probability of $1$ represents a sure or certain event. Any calculation yielding a negative number or a value greater than $1$ is mathematically invalid.
What are complementary events in Class 10 probability?
Complementary events are two mutually exclusive outcomes whose union encompasses the entire sample space. For any event $E$, the event ‘not $E$’ (denoted as $\bar{E}$ or $E’$) represents its complement. Their probabilities always sum to 1: $P(E) + P(\bar{E}) = 1$, allowing $P(\bar{E})$ to be calculated as $1 – P(E)$.
How many face cards are in a standard deck of 52 cards?
There are $12$ face cards (also called picture cards) in a standard deck of $52$ cards: $4$ Kings, $4$ Queens, and $4$ Jacks. Among these $12$ face cards, $6$ are red (from Hearts and Diamonds) and $6$ are black (from Spades and Clubs). Aces are not face cards.
What is an elementary event?
An elementary event is an event that consists of exactly one single outcome from the sample space. For example, when tossing a single coin, getting a Head ($H$) is an elementary event. The sum of the probabilities of all elementary events of any random experiment is always equal to $1$.
Why are the outcomes ‘two heads’, ‘two tails’, and ‘one of each’ not equally likely when tossing two coins?
When two coins are tossed, the sample space consists of four equally likely ordered pairs: $(H,H), (H,T), (T,H),$ and $(T,T)$. “Two heads” occurs once ($\frac{1}{4}$), “two tails” occurs once ($\frac{1}{4}$), but “one of each” occurs twice ($(H,T)$ and $(T,H)$), giving it a probability of $\frac{2}{4} = \frac{1}{2}$. Because their chances differ, the outcomes are not equally likely.
What is the sample space when three fair coins are tossed simultaneously?
When three coins are tossed simultaneously, there are $2^3 = 8$ equally likely outcomes: $S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}$. Each individual triple has an elementary probability of $\frac{1}{8}$.
How many outcomes are possible when rolling two dice together?
Rolling two dice simultaneously yields $6 \times 6 = 36$ equally likely ordered pair outcomes, ranging from $(1,1)$ to $(6,6)$. The sum of the numbers on the two dice ranges from a minimum of $2$ to a maximum of $12$.
What is the difference between mutually exclusive and independent events?
Mutually exclusive events cannot occur simultaneously in a single trial; the occurrence of one completely rules out the other ($P(A \cap B) = 0$). Independent events are events where the occurrence of one does not affect the probability of the other. In Class 10, the curriculum focuses on mutually exclusive, equally likely elementary events.
How is geometric probability calculated in Class 10 problems?
Geometric probability applies when outcomes correspond to points in a geometric region (length, area, or volume). The probability that a randomly chosen point falls into a specified sub-region is the ratio of the measure of the favourable sub-region to the measure of the total region: $P = \frac{\text{Favourable Area}}{\text{Total Area}}$.
Is 1 considered a prime number in dice probability questions?
No. The number $1$ is defined mathematically as neither prime nor composite. When rolling a standard 6-sided die, the prime numbers are strictly $2, 3,$ and $5$ (a total of $3$ prime outcomes), giving $P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$.
What is an impossible event and what is its probability?
An impossible event is an event that contains no favourable outcomes from the sample space ($n(E) = 0$) and cannot happen under any circumstances (e.g., rolling an $8$ on a standard die). The probability of an impossible event is always $0$.
What does ‘at least once’ mean in two-dice probability problems?
In probability, ‘at least once’ means one or more occurrences. For instance, the event “a 5 appears at least once” when rolling two dice includes outcomes with exactly one five ($(5,1), (1,5)$, etc.) as well as the outcome with two fives ($(5,5)$), giving $11$ favourable outcomes out of $36$.
What common mistake do students make in playing card probability questions?
A frequent mistake is confusing suits with colours. There are $2$ colours (Red and Black, with $26$ cards each) and $4$ suits (Spades, Clubs, Hearts, Diamonds, with $13$ cards each). Another common error is counting Aces as face cards; face cards refer only to Kings, Queens, and Jacks.
How should final probability answers be presented in CBSE board examinations?
Probabilities should be expressed as common fractions in lowest terms (e.g., $\frac{1}{4}$ rather than $\frac{9}{36}$) or as terminating/clean decimals. Unless explicitly instructed in the question, leaving results as simplified proper fractions is the accepted CBSE standard.