Ex 7.1 · Q1 Rank the following events on a scale from 0 (Impossible) to 1 (Certain): (i) Picking a card with a number between 2 and 10 from a standard deck, (ii) Rolling an 8 on a fair 6-sided die, (iii) Tossing a coin and getting heads, (iv) The sun rising in the east tomorrow, (v) Rolling a number less than 7 on a fair 6-sided die. [NCERT p.159]
Ex 7.1 · Q1 निम्नलिखित घटनाओं को 0 (Impossible) से 1 (Certain) के पैमाने पर रखें: (i) ताश की गड्डी से 2 से 10 के बीच का पत्ता निकालना, (ii) 6-sided die पर 8 आना, (iii) Coin toss पर heads आना, (iv) कल पूर्व में sunrise होना, (v) 6-sided die पर 7 से छोटी संख्या आना। [NCERT p.159]
1. Event Analysis
• (ii) Rolling an 8 on a 6-sided die: Faces are only $\{1, 2, 3, 4, 5, 6\} \implies P = \mathbf{0\text{ (Impossible)}}$.
• (iii) Tossing heads on fair coin: $P = \mathbf{0.5\text{ (Even chance)}}$.
• (i) Picking a numbered card 2 to 10: 9 ranks $\times$ 4 suits $= 36$ cards out of $52 \implies P = \frac{36}{52} = \frac{9}{13} \approx \mathbf{0.692\text{ (Likely)}}$.
• (iv) Sun rising in east tomorrow: Universal physical law $\implies P = \mathbf{1.0\text{ (Certain)}}$.
• (v) Rolling a number $< 7$ on a die: All 6 faces $\{1, 2, 3, 4, 5, 6\} < 7 \implies P = \frac{6}{6} = \mathbf{1.0\text{ (Certain)}}$.
🎯 Final Answer: Ranking from least to most likely: (ii) $0$ < (iii) $0.5$ < (i) $0.692$ < (iv) & (v) $1.0$.
1. मान विश्लेषण (ii) 8 आना $\implies P = \mathbf{0}$, (iii) Heads आना $\implies P = \mathbf{0.5}$, (i) 2 से 10 कार्ड $\implies P = \frac{36}{52} \approx \mathbf{0.692}$, (iv) पूर्व में sunrise $\implies P = \mathbf{1.0}$, (v) 7 से छोटी संख्या $\implies P = \mathbf{1.0}$।
🎯 final answer: order: (ii) $0$ < (iii) $0.5$ < (i) $0.692$ < (iv), (v) $1.0$।
5. Exercise Set 7.2 (NCERT pp. 165–166)
5. Exercise Set 7.2 (NCERT pp. 165–166)
Ex 7.2 · Q1 A teacher mixes a large bag of sweets of different colours and gives 50 students one sweet each at random. 15 students get red sweets, 20 get green, 10 get yellow, and 5 get blue. Find the experimental probability of getting: (i) a red sweet, (ii) a green sweet, (iii) a yellow or blue sweet. [NCERT p.165]
Ex 7.2 · Q1 एक शिक्षक अलग-अलग रंगों की मिठाइयों के बैग से 50 students को एक-एक sweets देते हैं। 15 को लाल, 20 को हरी, 10 को पीली और 5 को नीली sweets मिलती है। Experimental probability find करें: (i) लाल sweets, (ii) हरी sweets, (iii) पीली या नीली sweets। [NCERT p.165]
1. Total Trials $N = 50$.
2. Probabilities
• (i) $P(\text{Red}) = \frac{15}{50} = \frac{3}{10} = \mathbf{0.30\text{ (or } 30\%)}$.
• (ii) $P(\text{Green}) = \frac{20}{50} = \frac{2}{5} = \mathbf{0.40\text{ (or } 40\%)}$.
• (iii) $P(\text{Yellow or Blue}) = \frac{10 + 5}{50} = \frac{15}{50} = \frac{3}{10} = \mathbf{0.30\text{ (or } 30\%)}$.
🎯 Final Answer: (i) $\mathbf{0.30}$, (ii) $\mathbf{0.40}$, (iii) $\mathbf{0.30}$.
1. कुल प्रयास $N = 50$।
2. calculation (i) $P(\text{Red}) = \frac{15}{50} = \mathbf{0.30}$, (ii) $P(\text{Green}) = \frac{20}{50} = \mathbf{0.40}$, (iii) $P(\text{Yellow/Blue}) = \frac{15}{50} = \mathbf{0.30}$।
🎯 final answer: (i) $\mathbf{0.30}$, (ii) $\mathbf{0.40}$, (iii) $\mathbf{0.30}$।
Ex 7.2 · Q2 A survey is conducted at a school where a random sample of 40 students is asked about their favourite sport. 18 chose cricket, 12 chose football, 6 chose badminton, and 4 chose basketball. (i) Estimate the probability that a randomly selected student prefers cricket. (ii) If the school has 600 students, estimate how many prefer football. [NCERT p.165]
Ex 7.2 · Q2 40 students के survey में 18 क्रिकेट, 12 फुटबॉल, 6 बैडमिंटन और 4 बास्केटबॉल चुनते हैं। (i) क्रिकेट पसंद करने की estimated probability find करें। (ii) यदि स्कूल में 600 छात्र हैं, तो फुटबॉल पसंद करने वाले students की estimateित संख्या बताएं। [NCERT p.165]
1. Part (i) Cricket Probability
$$P(\text{Cricket}) = \frac{18}{40} = \frac{9}{20} = \mathbf{0.45\text{ (or } 45\%)}$$
2. Part (ii) Football Population Projection
$$P(\text{Football}) = \frac{12}{40} = 0.30 \implies \text{Expected Footballers} = 0.30 \times 600 = \mathbf{180\text{ students}}$$
🎯 Final Answer: (i) $\mathbf{0.45}$, (ii) $\mathbf{180\text{ students}}$.
1. भाग (i) Cricket $P(\text{Cricket}) = \frac{18}{40} = \mathbf{0.45}$।
2. भाग (ii) Football Projection $P(\text{Football}) = \frac{12}{40} = 0.30 \implies \text{students की संख्या} = 0.30 \times 600 = \mathbf{180}$।
🎯 final answer: (i) $\mathbf{0.45}$, (ii) $\mathbf{180\text{ छात्र}}$।
Ex 7.2 · Q4 Toss a paper cup into the air 50 times and observe whether it lands bottom-down, top-down, or on its side (Figure 7.5). If it lands on its side 38 times, bottom-down 8 times, and top-down 4 times, find the experimental probability for each orientation. Why can theoretical probability not be easily computed here? [NCERT p.165]
Ex 7.2 · Q4 एक paper cup को 50 बार हवा में उछालने पर यदि वह 38 बार side पर, 8 बार bottom पर और 4 बार top पर गिरता है, तो हर position की experimental probability find करें। यहां theoretical probability आसानी से क्यों नहीं निकाली जा सकती? [NCERT p.165]
1. Experimental Probabilities
• $P(\text{Side}) = \frac{38}{50} = \mathbf{0.76}$
• $P(\text{Bottom}) = \frac{8}{50} = \mathbf{0.16}$
• $P(\text{Top}) = \frac{4}{50} = \mathbf{0.08}$
2. Why Theoretical Fails A paper cup is an asymmetric 3D object with non-uniform mass distribution, open rim aerodynamics, and unequal contact surfaces. Outcomes are not equally likely, requiring empirical observation rather than classical geometric symmetry.
🎯 Final Answer: Side: $\mathbf{0.76}$, Bottom: $\mathbf{0.16}$, Top: $\mathbf{0.08}$. Theoretical calculation is impractical due to geometric asymmetry.
1. Experimental calculation Side: $\frac{38}{50} = \mathbf{0.76}$, Bottom: $\frac{8}{50} = \mathbf{0.16}$, Top: $\frac{4}{50} = \mathbf{0.08}$।
2. Theoretical calculation का कारण Paper cup की आकृति asymmetric होने के कारण outcomes equally likely नहीं होते।
🎯 final answer: Side: $\mathbf{0.76}$, Bottom: $\mathbf{0.16}$, Top: $\mathbf{0.08}$।
Ex 7.2 · Q6 Suppose you roll a 6-sided die 12 times and get a '3' three times. (i) What is the experimental probability of getting a '3'? (ii) What is the theoretical probability of getting a '3'? (iii) Explain why they differ and how they can be brought closer. [NCERT p.166]
Ex 7.2 · Q6 एक 6-sided die को 12 बार फेंकने पर 3 बार '3' आता है। (i) '3' आने की experimental probability क्या है? (ii) Theoretical probability क्या है? (iii) दोनों में अंतर क्यों है और इन्हें पास कैसे लाया जा सकता है? [NCERT p.166]
1. Experimental vs Theoretical
• (i) $P_{\text{exp}}(3) = \frac{3}{12} = \frac{1}{4} = \mathbf{0.25}$.
• (ii) $P_{\text{theo}}(3) = \frac{1}{6} \approx \mathbf{0.167}$.
2. Explanation & Law of Large Numbers In a small sample ($N=12$), random statistical fluctuations dominate. By increasing the number of trials to $N = 1000$ or $10000$, the Law of Large Numbers ensures $P_{\text{exp}}(3)$ converges tightly to $\frac{1}{6}$.
🎯 Final Answer: Experimental $= \mathbf{0.25}$, Theoretical $= \mathbf{0.167}$. Increasing trials to hundreds or thousands eliminates sample variation.
1. दोनों का मान (i) $P_{\text{exp}}(3) = \frac{3}{12} = \mathbf{0.25}$, (ii) $P_{\text{theo}}(3) = \frac{1}{6} \approx \mathbf{0.167}$।
2. अंतर का कारण 12 का sample size बहुत छोटा है। जब rolls की संख्या $1000$ या $10000$ की जाएगी तो empirical probability theoretical मान $\frac{1}{6}$ के करीब पहुंच जाएगी।
🎯 final answer: Experimental: $\mathbf{0.25}$, Theoretical: $\mathbf{0.167}$।
8. End-of-Chapter Exercises (NCERT pp. 169–173)
8. End-of-Chapter Exercises (NCERT pp. 169–173)
End Ex · Q1 Fill in the blanks: (i) The probability of an impossible event is _______, (ii) The set of all possible outcomes of a random experiment is called the _______, (iii) The probability of an event that is certain to happen is _______, (iv) Tossing a fair coin has a probability of _______ for getting heads. [NCERT pp. 169–170]
End Ex · Q1 रिक्त स्थान भरें: (i) Impossible event की probability _______ होती है, (ii) Random experiment के सभी possible outcomesों का समुच्चय _______ कहलाता है, (iii) Certain event की probability _______ होती है, (iv) Fair coin toss पर heads आने की probability _______ होती है। [NCERT pp. 169–170]
1. Fundamental Definitions
• (i) Impossible event: $\mathbf{0}$
• (ii) Set of all outcomes: Sample Space ($S$)
• (iii) Certain event: $\mathbf{1}$
• (iv) Heads on fair coin: $\mathbf{\frac{1}{2}\text{ (or } 0.5)}$
🎯 Final Answer: (i) 0, (ii) Sample Space, (iii) 1, (iv) $\frac{1}{2}$.
1. answer (i) 0, (ii) Sample Space ($S$), (iii) 1, (iv) $\frac{1}{2}$।
🎯 final answer: (i) 0, (ii) Sample Space, (iii) 1, (iv) $\frac{1}{2}$।
End Ex · Q2 & Q3 (i) In a survey of 50 students, 15 said they liked football. The frequency is 15 and relative frequency is _______. (ii) Which experiments have equally likely outcomes: Starting a car, Tossing a fair coin, Rolling a fair die, Drawing a marble from 3 red & 7 blue marbles, A newborn baby is boy or girl? [NCERT p.170]
End Ex · Q2 & Q3 (i) 50 students में से 15 फुटबॉल पसंद करते हैं। Relative frequency _______ होगी। (ii) कौन से experiment equally likely हैं: कार स्टार्ट करना, Coin toss, Die roll, 3 लाल व 7 नीली गोलियों में से निकालना, नवजात शिशु लड़का या लड़की होना? [NCERT p.170]
1. Q2 Relative Frequency $\text{Relative Frequency} = \frac{15}{50} = \mathbf{0.30\text{ (or } \frac{3}{10})}$.
2. Q3 Equally Likely Analysis
• (i) Car starts: Not equally likely (depends on mechanical condition).
• (ii) Fair coin toss: Equally likely ($P(H) = P(T) = 0.5$).
• (iii) Fair 6-sided die: Equally likely (each face $P = 1/6$).
• (iv) 3 red & 7 blue marbles: Not equally likely ($P(R) = 0.3, P(B) = 0.7$).
• (v) Newborn baby boy/girl: Equally likely ($P(B) = P(G) = 0.5$).
🎯 Final Answer: Q2: $\mathbf{0.30}$; Q3 Equally likely: (ii), (iii), and (v).
1. Q2 Relative Frequency $\text{Relative Frequency} = \frac{15}{50} = \mathbf{0.30}$।
2. Q3 विश्लेषण (ii) Fair coin, (iii) Fair die, और (v) Baby boy/girl equally likely हैं।
🎯 final answer: Q2: $\mathbf{0.30}$; Q3: (ii), (iii), और (v)।
End Ex · Q4 Calculate the probabilities: (i) Two coins tossed: $P(\text{at least 1 head})$, (ii) Card numbered 1 to 10: $P(\text{even number})$, (iii) Die rolled: $P(\text{number } > 4)$, (iv) Bag of 3 red, 2 blue, 1 green: $P(\text{not red})$, (v) Three coins tossed: $P(\text{exactly 2 heads})$. [NCERT p.170]
End Ex · Q4 Probability find करें: (i) 2 coins: $P(\text{कम से कम 1 head})$, (ii) 1 से 10 कार्ड: $P(\text{even number})$, (iii) Die: $P(\text{संख्या } > 4)$, (iv) 3 लाल, 2 नीली, 1 हरी: $P(\text{लाल नहीं})$, (v) 3 coins: $P(\text{ठीक 2 heads})$। [NCERT p.170]
1. Calculations (i) to (iii)
• (i) $S = \{HH, HT, TH, TT\}$, at least 1 head $= \{HH, HT, TH\} \implies P = \mathbf{\frac{3}{4}}$.
• (ii) Even in $\{1..10\} = \{2, 4, 6, 8, 10\} \implies P = \frac{5}{10} = \mathbf{\frac{1}{2}}$.
• (iii) Number $> 4$ on die $= \{5, 6\} \implies P = \frac{2}{6} = \mathbf{\frac{1}{3}}$.
2. Calculations (iv) to (v)
• (iv) Total balls $= 6$; Not red $= 2 + 1 = 3 \implies P = \frac{3}{6} = \mathbf{\frac{1}{2}}$.
• (v) Exactly 2 heads in 3 coins $= \{HHT, HTH, THH\} \implies P = \mathbf{\frac{3}{8}}$.
🎯 Final Answer: (i) $\mathbf{\frac{3}{4}}$, (ii) $\mathbf{\frac{1}{2}}$, (iii) $\mathbf{\frac{1}{3}}$, (iv) $\mathbf{\frac{1}{2}}$, (v) $\mathbf{\frac{3}{8}}$.
1. भाग (i) से (iii) (i) $\mathbf{\frac{3}{4}}$, (ii) $\mathbf{\frac{1}{2}}$, (iii) $\mathbf{\frac{1}{3}}$।
2. भाग (iv) से (v) (iv) $\mathbf{\frac{1}{2}}$, (v) $\mathbf{\frac{3}{8}}$।
🎯 final answer: (i) $\frac{3}{4}$, (ii) $\frac{1}{2}$, (iii) $\frac{1}{3}$, (iv) $\frac{1}{2}$, (v) $\frac{3}{8}$।
End Ex · Q6 A child has 2 shirts (red $R$, blue $B$) and 3 types of pants (jeans $J$, khakis $K$, shorts $S$). List all possible combinations in a table format and find $n(S)$. [NCERT p.171]
End Ex · Q6 एक बच्चे के पास 2 शर्ट (लाल $R$, नीली $B$) और 3 पैंट (जीन्स $J$, खाकी $K$, शॉर्ट्स $S$) हैं। सभी possible संयोजनों की तालिका बनाएं और $n(S)$ find करें। [NCERT p.171]
1. Combination Matrix
• Red Shirt: $(R, J), (R, K), (R, S)$ (3 outfits)
• Blue Shirt: $(B, J), (B, K), (B, S)$ (3 outfits)
2. Total Sample Space Size
$$n(S) = 2 \text{ shirts} \times 3 \text{ pants} = \mathbf{6\text{ possible outfits}}$$
🎯 Final Answer: $n(S) = \mathbf{6\text{ distinct outfits}}$.
1. तालिका लाल शर्ट के 3 संयोजन और नीली शर्ट के 3 संयोजन बनते हैं।
2. कुल संयोजन $n(S) = 2 \times 3 = \mathbf{6}$।
🎯 final answer: कुल $n(S) = \mathbf{6}$ संयोजन बनते हैं।
End Ex · Q7 A tyre company records distances before replacement in 1000 cases: <4000 km (20), 4001–9000 km (210), 9001–14000 km (325), >14000 km (445). Find the probability that a tyre lasts: (i) <4000 km, (ii) between 4000 and 14000 km, (iii) >14000 km. [NCERT p.171]
End Ex · Q7 टायर कंपनी के 1000 मामलों में दूरी: <4000 km (20), 4001–9000 km (210), 9001–14000 km (325), >14000 km (445)। Probability find करें: (i) <4000 km, (ii) 4000 से 14000 km, (iii) >14000 km। [NCERT p.171]
1. Probabilities ($N=1000$)
• (i) $P(< 4000) = \frac{20}{1000} = \mathbf{0.02}$
• (ii) $P(4000\text{ to }14000) = \frac{210 + 325}{1000} = \frac{535}{1000} = \mathbf{0.535}$
• (iii) $P(> 14000) = \frac{445}{1000} = \mathbf{0.445}$
🎯 Final Answer: (i) $\mathbf{0.02}$, (ii) $\mathbf{0.535}$, (iii) $\mathbf{0.445}$.
1. calculation (i) $\frac{20}{1000} = \mathbf{0.02}$, (ii) $\frac{535}{1000} = \mathbf{0.535}$, (iii) $\frac{445}{1000} = \mathbf{0.445}$।
🎯 final answer: (i) $\mathbf{0.02}$, (ii) $\mathbf{0.535}$, (iii) $\mathbf{0.445}$।
End Ex · Q8 The letters of the word 'PEACE' are placed on cards. Leela draws a card without looking. (i) What is the probability that it is P, E, or C? (ii) What is the probability that it is NOT an E? [NCERT p.171]
End Ex · Q8 शब्द 'PEACE' के अक्षर कार्डों पर लिखे हैं। लीला एक कार्ड निकालती है। (i) P, E या C आने की probability क्या है? (ii) E न आने की probability क्या है? [NCERT p.171]
1. Sample Space Letters are $\{P, E_1, A, C, E_2\} \implies n(S) = 5$.
2. Part (i) P, E, or C Favourable letters are $\{P, E_1, C, E_2\}$ (4 letters) $\implies P = \mathbf{\frac{4}{5} = 0.8}$.
3. Part (ii) Not an E Favourable letters are $\{P, A, C\}$ (3 letters) $\implies P = \mathbf{\frac{3}{5} = 0.6}$.
🎯 Final Answer: (i) $\mathbf{\frac{4}{5} = 0.8}$, (ii) $\mathbf{\frac{3}{5} = 0.6}$.
1. Sample Space $n(S) = 5$ अक्षर हैं।
2. calculation (i) $\{P, E, C, E\} \implies P = \mathbf{\frac{4}{5}}$, (ii) $\{P, A, C\} \implies P = \mathbf{\frac{3}{5}}$।
🎯 final answer: (i) $\mathbf{\frac{4}{5}}$, (ii) $\mathbf{\frac{3}{5}}$।
End Ex · Q9 A spinning arrow comes to rest pointing at one of the numbers $1, 2, 3, 4, 5, 6, 7, 8$ with equal likelihood (Figure 7.7). Find the probability that it points at: (i) 8, (ii) an odd number, (iii) a number greater than 2, (iv) a number less than 9, (v) a multiple of 3. [NCERT p.171]
End Ex · Q9 एक घूमने वाला तीर 1 से 8 तक के किसी एक अंक पर रुकता है (Figure 7.7)। Probability find करें कि यह रुकेगा: (i) 8 पर, (ii) विषम संख्या पर, (iii) 2 से बड़ी संख्या पर, (iv) 9 से छोटी संख्या पर, (v) 3 के गुणज पर। [NCERT p.171]
1. Sample Space $S = \{1, 2, 3, 4, 5, 6, 7, 8\} \implies n(S) = 8$.
2. Calculations (i) to (v)
• (i) Point at 8: $\{8\} \implies P = \mathbf{\frac{1}{8}}$.
• (ii) Odd number: $\{1, 3, 5, 7\} \implies P = \frac{4}{8} = \mathbf{\frac{1}{2}}$.
• (iii) Number $> 2$: $\{3, 4, 5, 6, 7, 8\} \implies P = \frac{6}{8} = \mathbf{\frac{3}{4}}$.
• (iv) Number $< 9$: All 8 numbers $\implies P = \frac{8}{8} = \mathbf{1\text{ (Certain)}}$.
• (v) Multiple of 3: $\{3, 6\} \implies P = \frac{2}{8} = \mathbf{\frac{1}{4}}$.
🎯 Final Answer: (i) $\mathbf{\frac{1}{8}}$, (ii) $\mathbf{\frac{1}{2}}$, (iii) $\mathbf{\frac{3}{4}}$, (iv) $\mathbf{1}$, (v) $\mathbf{\frac{1}{4}}$.
1. Sample Space $n(S) = 8$।
2. calculation (i) $\mathbf{\frac{1}{8}}$, (ii) $\mathbf{\frac{1}{2}}$, (iii) $\mathbf{\frac{3}{4}}$, (iv) $\mathbf{1}$, (v) $\mathbf{\frac{1}{4}}$।
🎯 final answer: (i) $\frac{1}{8}$, (ii) $\frac{1}{2}$, (iii) $\frac{3}{4}$, (iv) $1$, (v) $\frac{1}{4}$।
End Ex · Q10 A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside (without replacement), and a second ball is drawn. Draw a tree diagram and find: (i) $P(\text{Red then Blue})$, (ii) $P(\text{Both Blue})$. [NCERT pp. 171–172]
End Ex · Q10 एक basket में 4 लाल और 5 नीली गेंदें हैं। एक गेंद निकालकर अलग रख दी जाती है (without replacement), फिर दूसरी गेंद निकाली जाती है। Tree diagram बनाकर find करें: (i) $P(\text{Red then Blue})$, (ii) $P(\text{Both Blue})$। [NCERT pp. 171–172]
1. Multi-Stage Probabilities
• Total initially $= 9$. Draw 1: $P(R) = \frac{4}{9}, P(B) = \frac{5}{9}$.
• Draw 2 after Red: 8 balls remain (3 R, 5 B) $\implies P(B|R) = \frac{5}{8}$.
• Draw 2 after Blue: 8 balls remain (4 R, 4 B) $\implies P(B|B) = \frac{4}{8} = \frac{1}{2}$.
2. Path Probabilities
• (i) $P(R \text{ then } B) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \mathbf{\frac{5}{18} \approx 0.278}$.
• (ii) $P(B \text{ and } B) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \mathbf{\frac{5}{18} \approx 0.278}$.
🎯 Final Answer: (i) $\mathbf{\frac{5}{18}}$, (ii) $\mathbf{\frac{5}{18}}$.
1. Tree Diagram मान पहली गेंद के बाद 8 गेंदें बचती हैं।
2. calculation (i) $P(RB) = \frac{4}{9} \times \frac{5}{8} = \mathbf{\frac{5}{18}}$, (ii) $P(BB) = \frac{5}{9} \times \frac{4}{8} = \mathbf{\frac{5}{18}}$।
🎯 final answer: (i) $\mathbf{\frac{5}{18}}$, (ii) $\mathbf{\frac{5}{18}}$।
End Ex · Q11 & Q12 (i) When throwing a pair of dice, give an impossible event ($P=0$) and a sure event ($P=1$). (ii) Find the probability that the sum of numbers on two rolled dice is a prime number greater than 5. [NCERT p.172]
End Ex · Q11 & Q12 (i) दो पासे फेंकने पर impossible event ($P=0$) और sure event ($P=1$) का example दें। (ii) दो पासों के योगफल के 5 से बड़े prime number होने की probability find करें। [NCERT p.172]
1. Q11 Impossible and Sure Events
• Impossible ($P=0$): 'Getting a sum of 13' (Max sum is $6+6=12$).
• Sure event ($P=1$): 'Getting a sum between 2 and 12 inclusive'.
2. Q12 Prime Sum $> 5$ Total pairs $n(S) = 36$. Primes $> 5$ up to 12 are $7$ and $11$:
• Sum $= 7$: $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$ (6 pairs)
• Sum $= 11$: $\{(5,6), (6,5)\}$ (2 pairs)
• Total favourable $= 6 + 2 = 8 \implies P = \frac{8}{36} = \mathbf{\frac{2}{9} \approx 0.222}$.
🎯 Final Answer: Q11: Impossible = 'Sum 13', Sure = 'Sum $\le 12$'; Q12: $P = \mathbf{\frac{2}{9}}$.
1. Q11 example Impossible: 'योग 13 आना' ($P=0$), Sure: 'योग 12 या उससे कम होना' ($P=1$)।
2. Q12 calculation Prime sum $> 5$ के लिए $7$ और $11$ के कुल 8 अनुकूल outcomes हैं $\implies P = \frac{8}{36} = \mathbf{\frac{2}{9}}$।
🎯 final answer: Q11: Sum 13 ($P=0$), Sum $\le 12$ ($P=1$); Q12: $\mathbf{\frac{2}{9}}$।
End Ex · Q13 & Q14 (i) Compare sample space sizes for drawing 2 balls from $\{1, 2, 3, 4\}$ with replacement vs without replacement. (ii) List the sample space for simultaneously tossing a coin and drawing a card from $\{1, 2, 3, 4, 5, 6\}$. [NCERT p.172]
End Ex · Q13 & Q14 (i) $\{1, 2, 3, 4\}$ से 2 गेंदें निकालने पर with replacement और without replacement के sample space sizes की तुलना करें। (ii) Coin toss और 1 से 6 कार्ड निकालने का sample space लिखें। [NCERT p.172]
1. Q13 Sample Space Sizes
• With replacement: $4 \times 4 = \mathbf{16\text{ pairs}}$ (e.g. $(1,1), (1,2), \dots$)
• Without replacement: $4 \times 3 = \mathbf{12\text{ pairs}}$ (diagonal pairs $(k,k)$ excluded).
2. Q14 Coin + Card $S$
$$S = \{(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\} \implies n(S) = \mathbf{12}$$
🎯 Final Answer: Q13: With replacement $= \mathbf{16}$, Without replacement $= \mathbf{12}$; Q14: $n(S) = \mathbf{12}$.
1. Q13 Sizes With replacement $= 4 \times 4 = \mathbf{16}$, Without replacement $= 4 \times 3 = \mathbf{12}$।
2. Q14 Sample Space $n(S) = 2 \times 6 = \mathbf{12}$ जोड़े बनते हैं।
🎯 final answer: Q13: 16 और 12; Q14: $n(S) = 12$।
End Ex · Q15 Three coins are tossed, and the number of heads is recorded. Which of the following lists is a valid sample space: (i) $\{1, 2, 3\}$, (ii) $\{0, 1, 2\}$, (iii) $\{0, 1, 2, 3, 4\}$, (iv) $\{0, 1, 2, 3\}$? Why do the others fail? [NCERT pp. 172–173]
End Ex · Q15 3 सिक्के उछालकर heads की संख्या दर्ज की जाती है। इनमें से कौन सा सही sample space है: (i) $\{1, 2, 3\}$, (ii) $\{0, 1, 2\}$, (iii) $\{0, 1, 2, 3, 4\}$, (iv) $\{0, 1, 2, 3\}$? अन्य विकल्प क्यों गलत हैं? [NCERT pp. 172–173]
1. Exhaustiveness Criterion A valid sample space must cover all mutually exclusive and exhaustive possibilities. In 3 coin tosses, heads can be $0$ (TTT), $1$ (HTT, THT, TTH), $2$ (HHT, HTH, THH), or $3$ (HHH).
2. Evaluation
• (i) Missing outcome $0$ heads (not exhaustive).
• (ii) Missing outcome $3$ heads (not exhaustive).
• (iii) Includes $4$ heads, which is impossible with only 3 coins.
• (iv) $\{0, 1, 2, 3\}$ is strictly valid and exhaustive.
🎯 Final Answer: (iv) $\{0, 1, 2, 3\}$ is the only valid sample space.
1. Sample Space rule Sample space में सभी possible outcomes होने चाहिए। 3 सिक्कों में heads की संख्या $0, 1, 2, 3$ हो सकती है।
2. मूल्यांकन (i) में 0 गायब है, (ii) में 3 गायब है, (iii) में अpossible 4 शामिल है। केवल (iv) सही है।
🎯 final answer: केवल (iv) $\{0, 1, 2, 3\}$ सही sample space है।
End Ex · Q16 A die is dropped at random on a rectangular region of length $3\text{ m}$ and width $2\text{ m}$ containing a circular region of diameter $1\text{ m}$ (Figure 7.8). What is the geometric probability that it will land inside the circle? [NCERT p.173]
End Ex · Q16 $3\text{ m} \times 2\text{ m}$ आयताकार region जिसके अंदर $1\text{ m}$ व्यास का वृत्ताकार region बना है (Figure 7.8), पर एक पासा गिराया जाता है। पासे के वृत्त के अंदर गिरने की geometric probability find करें। [NCERT p.173]
1. Total Bounding Area
$$\text{Area of Rectangle } A_{\text{total}} = 3\text{ m} \times 2\text{ m} = \mathbf{6\text{ m}^2}$$
2. Circular Target Area Diameter $d = 1\text{ m} \implies r = 0.5\text{ m} = \frac{1}{2}\text{ m}$:
$$A_{\text{circle}} = \pi r^2 = \pi \left(\frac{1}{2}\right)^2 = \mathbf{\frac{\pi}{4}\text{ m}^2}$$
3. Geometric Probability
$$P(\text{Lands in circle}) = \frac{A_{\text{circle}}}{A_{\text{total}}} = \frac{\frac{\pi}{4}}{6} = \mathbf{\frac{\pi}{24} \approx 0.1309 \text{ (or } 13.09\%)}$$
🎯 Final Answer: Probability $= \mathbf{\frac{\pi}{24} \approx 0.131}$.
1. कुल Area Rectangle का Area $= 3 \times 2 = \mathbf{6\text{ m}^2}$।
2. Circle Area Circle का Area $= \pi (0.5)^2 = \mathbf{\frac{\pi}{4}\text{ m}^2}$।
3. Geometric Probability $P = \frac{\pi/4}{6} = \mathbf{\frac{\pi}{24} \approx 0.131}$।
🎯 final answer: Probability $= \mathbf{\frac{\pi}{24} \approx 0.131}$ है।