Probability: Class 10 Maths Chapter 14 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

70 min
3 simulations
33 practice questions
25 MCQs

Is chapter mein aap kya seekhenge

Tay karo ki kisi experiment ke outcomes equally likely hain ya nahi, aur samjhao ki theoretical probability is assumption par kyun tiki hai.
Kisi event ki theoretical probability nikaalo, favourable outcomes ko sabhi outcomes se divide karke: coins, dice, cards, spinners aur bag se draw ke liye.
Elementary events pehchano aur is fact ka use karo ki unki probabilities ka total 1 hota hai.
Complement rule se "not E" ki probability nikaalo, aur event ya uske complement mein se jise ginna aasaan ho, use chuno.
Impossible aur sure events pehchano aur samjhao ki har probability 0 se 1 ke beech hi kyun hoti hai.
Do coins aur do dice ke ordered outcomes list karo, without replacement wale draws handle karo, aur un arguments ko pakdo jo unequal results ko equally likely maan lete hain.
Jab outcomes gine na ja sakein, tab probability ko lengths aur areas tak le jao (board syllabus se aage).

NCERT Mathematics Class 10, Chapter 14, pp. 202–217 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

14.1 Probability — Ek Theoretical Approach

Gully-cricket match se pehle ek captain coin uchhaalta hai aur doosra call karta hai. Koi nahi bata sakta ki kaunsa face upar aayega, phir bhi dono captains toss ko fair maante hain. Kyun? Kyunki har face ka chance same hai. Class 9 mein tumne experiment ko kai baar dohra kar aur results gin kar chances ka andaaza lagaya tha. Is chapter mein tum chances ko bilkul exact calculate karoge, possible outcomes ke baare mein soch kar, bas ek condition honi chahiye: outcomes equally likely hone chahiye.

Equally likely outcomes

Coin heads (H) ya tails (T) dikhata hai. In dono results ka chance kab bilkul same hota hai? Do baatein sach honi chahiye. Coin fair hona chahiye: barabar bana hua, koi side bhaari nahi aur aisi shape nahi ki ek side zyada baar upar aaye. Aur toss at random (यादृच्छया) hona chahiye: coin freely ghoome aur gire, aur koi use control na kare. Tab H aur T equally likely (समप्रायिक) outcomes hote hain.

Die bhi aisa hi behave karta hai. Agar uske chhe faces mein sirf dots ka fark hai, to 1 se 6 tak koi bhi number favoured nahi hota. Jab tak question kuch aur na kahe, is chapter ka har coin aur har die fair hai.

Definition

Equally likely outcomes

Kisi experiment ke outcomes equally likely tab hote hain jab har outcome ke hone ka chance same ho, kyunki set-up mein koi cheez kisi ek outcome ka favour nahi karti.

ExampleFair die ke chhe faces; achhi tarah shuffle kiye deck ke 52 cards.

Lekin dhyaan rakho. Jin results ki humein parwaah hai, woh hamesha equally likely nahi hote, bhale hi objects equally likely hon. School mela mein ek bowl mein chhe ek jaisi mudi hui chits hain: paanch par Try again likha hai aur ek par Prize. Kya jeetna utna hi likely hai jitna na jeetna? Fig. 14.1 dekho.

3/3Try again: $\frac{5}{6} .$ Prize: $\frac{1}{6} .$ Dono results equally likely nahi hain.

Fig. 14.1Chhe chits equally likely hain; do results nahi.NCERT p. 203

1 se 6 number wali chhe ek jaisi mudi hui paper chits ek line mein khadi hain. Pehle step mein woh band hain, aur neeche ek line kehti hai ki kisi bhi ek chit ki probability one-sixth hai. Doosre step mein chits khulti hain: chit 1 se 5 par cross hai, yaani try again, aur chit 6 par star hai, yaani prize; braces paanch try-again chits aur akeli prize chit ko group karte hain. Teesre step mein do bars dono results compare karte hain: try-again bar prize bar se paanch guna lamba hai, aur un par five-sixths aur one-sixth likha hai.

Isliye outcomes list karne ka safe tareeka hai chit-by-chit, result-by-result nahi. Prize aur try again equally likely nahi hain, lekin chhe chits mein se har ek hai. Yahi trap tab bhi aata hai jab objects unequal numbers mein hon, jaise kisi bag mein ek colour ki balls doosre colour se zyada hon. Is chapter ki baaki har problem ek baat maan kar chalti hai: jo outcomes hum list karte hain, woh equally likely hain.

SimulationDo colours, unequal sectors

Ek spinner jiska red sector wheel ka teen-chauthai hissa ghere hai aur blue sector baaki ek-chauthai. Tum ek press mein ise 1, 10, 100 ya 1000 baar spin kar sakte ho; bars dikhate hain ki har colour kitni baar aaya, saath mein three-quarters aur one-quarter ke theoretical shares bhi.

Fig. 14.2Ek wheel jiska teen-chauthai hissa red hai. Kya red aur blue equally likely hain?

Try karo

Spin karne se pehle: kya red ka chance $\frac{1}{2}$ hai? 100 baar spin karo aur check karo.
Socho ki red hissa teen barabar sectors mein kata hai. Tab wheel mein kitne barabar sectors hain, aur unmein se kitne red hain?

Theoretical probability

Class 9 mein tumne experimental probability dekhi thi: ek experiment kai baar dohrao, gino ki ek event $E$ kitni baar hota hai, aur trials ki sankhya se divide karo.

Experimental probability

P (E) = \frac{f}{N}

(14.1)

jahan

$f$: un trials ki sankhya jinmein $E$ hua
$N$: trials ki total sankhya

Kya engineers ek naye dam ke fail hone ka chance sau floods ka intezaar karke judge kar sakte hain? Kya koi airline sau naye aircraft crash karke dekh sakti hai ki wing kitni baar tootta hai? Bilkul nahi. Coins aur dice ke liye trials dohrana sasta hai, lekin kai real questions ke liye yeh bahut mehnga, khatarnaak ya slow hai. Humein aisa tareeka chahiye jisse kuch bhi dohraye bina probability mil jaaye.

Raasta ek assumption hai. Agar hum dikha sakein ki outcomes equally likely hain, to humein koi data nahi chahiye: bas ginna hai. Sabhi possible outcomes mein se jo event ko hone dete hain, unhe favourable outcomes (अनुकूल परिणाम) kehte hain.

Definition

Theoretical probability

Jab kisi experiment ke sabhi outcomes equally likely hon, to event $E$ ki theoretical (ya classical) probability hai $E$ ke favourable outcomes ki sankhya divided by sabhi possible outcomes ki sankhya.

ExampleMela bowl se ek chit: $P (prize) = \frac{1}{6} .$

Theoretical probability

P (E) = \frac{m}{n}

(14.2)

jahan

$m$: $E$ ke favourable outcomes ki sankhya
$n$: sabhi possible outcomes ki sankhya, sab equally likely

Jab bhi yeh chapter sirf probability kahe, to matlab yahi theoretical wali hai.

Quick check

3 mein se 1

Kis experiment mein list kiye gaye outcomes equally likely hain?

Ek batter ek ball face karta hai: out ya not outKal Shillong mein: baarish ya baarish nahi1 se 10 number wali 10 ek jaisi chits ka achhi tarah mila bowl: koi bhi ek number7 green aur 3 white marbles ka bag: green ya white

Answer dekho

Answer (C) 1 se 10 number wali 10 ek jaisi chits ka achhi tarah mila bowl: koi bhi ek number

Ek jaisi, achhi tarah mili chits har number ko same chance deti hain. Batter ki skill, mausam aur marbles ka 7-to-3 split, sab ek result ka favour karte hain, isliye woh pairs equally likely nahi hain.

Elementary events

Chalo definition ko kaam mein lagate hain, shuru karte hain aise events se jinmein sirf ek outcome ho.

Example 14.1NCERT Example 1 · p. 204

Toss ka call

Match se pehle visiting captain heads call karta hai. Coin fair hai. Call sahi hone ki probability kya hai? Galat hone ki probability kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14.2NCERT Example 2 · p. 204

Pencil box mein chaar pens

Ek pencil box mein chaar pens hain jo dekhne aur chhoone mein ek jaise hain, ek-ek red, green, blue aur black ink wala. Aditi aankhein band karke ek pen nikaalti hai. Probability nikaalo ki woh (a) green, (b) black, (c) red, (d) blue mein likhta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Definition

Elementary event

Aisa event jismein experiment ka bilkul ek outcome ho.

ExampleDie ke ek roll par 5 aana; Aditi ka green pen nikaalna.

Example 14.1 mein events sahi aur galat elementary hain, aur Example 14.2 ke chaar colours bhi. Ab unki probabilities jodo: $\frac{1}{2} + \frac{1}{2} = 1$ aur $\frac{1}{4} \times 4 = 1.$ Yeh koi sanyog nahi hai.

PropertyElementary events ka total 1 hota hai

Kisi bhi experiment ke sabhi elementary events ki probabilities ka total 1 hota hai.

Jin events ki humein parwaah hoti hai, unmein se zyadatar mein kai outcomes hote hain. Tab hum un sabko ginte hain.

Example 14.3NCERT Example 3 · p. 205

Ek draw ke do events

Ek bowl mein 1 se 10 number wali 10 ek jaisi chits hain, aur ek chit at random nikaali jaati hai. Probability nikaalo ki uska number (a) 4 se kam hai, (b) 4 ya zyada hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14.3 ka koi bhi event elementary nahi hai: ek mein teen outcomes hain aur doosre mein saat. Phir bhi unki probabilities, $\frac{3}{10}$ aur $\frac{7}{10},$ ka total phir se 1 hai. Agla section batata hai kyun.

Quick check

3 mein se 1

Ek die ek baar roll kiya jaata hai. Inmein se kaunsa elementary event hai?

even number aana3 se chhota number aanaprime number aana4 aana

Answer dekho

Answer (D) 4 aana

Sirf 4 aana mein bilkul ek outcome hai. 3 se chhota mein do outcomes hain (1 aur 2), aur even aur prime mein teen-teen.

Complementary events

Example 14.3 mein 4 se kam aur 4 ya zyada das chits ko aapas mein baant lete hain: har chit do events mein se bilkul ek mein aati hai. Jab bhi ek hota hai, doosra nahi hota. Aisi jodi ka apna naam hai.

Definition

Event ka complement

Maan lo ek event hai $E,$ tab event not $E,$ jise likhte hain $\overset{ˉ}{E},$ theek tab hota hai jab $E$ nahi hota. $E$ aur $\overset{ˉ}{E}$ ko complementary events kehte hain.

ExampleAgar chit par $E$ 4 se kam number hai, to $\overset{ˉ}{E}$ 4 ya usse zyada number hai.

$P (E)$ aur $P (\overset{ˉ}{E})$ ka total 1 kyun hona chahiye? Maan lo ek experiment ke $n$ equally likely outcomes hain, aur unmein se $m$ outcomes is event ke favourable hain: $E .$ Baaki har outcome par jo event hota hai woh hai $\overset{ˉ}{E},$ isliye $\overset{ˉ}{E}$ ke $n - m$ favourable outcomes hain aur $P (\overset{ˉ}{E}) = \frac{n - m}{n} .$ Jodne par $P (E) + P (\overset{ˉ}{E}) = \frac{n}{n} = 1.$ Ise rearrange karo to yahi complement rule hai.

Complement rule

P (\overset{ˉ}{E}) = 1 - P (E)

(14.3)

Fig. 14.3 yeh idea 12 bangles ke box se dikhata hai, jinmein 5 glass ki hain aur 7 metal ki, aur unmein se ek at random uthayi jaati hai.

4/4 $\frac{5}{12} + \frac{7}{12} = 1 :$ dono events bar ko bhar dete hain.

Fig. 14.3Event aur uska complement sabhi outcomes aapas mein baant lete hain, isliye dono milkar bar ko 0 se 1 tak bhar dete hain.NCERT p. 206

Barah bangles do rows mein rakhi hain: paanch patli, double-ring wali glass bangles aur saat moti metal bangles, neeche ek key ke saath. Pehle step mein panel kehta hai ki 12 bangles mein se ek at random uthao, isliye n barabar 12. Doosre step mein ek dashed outline paanch glass bangles ko gherti hai, yeh event E hai, panel P of E five-twelfths batata hai, aur 0 se 1 wale bar ka left hissa bhar jaata hai. Teesre step mein outline saat metal bangles par chali jaati hai, yaani not E, probability seven-twelfths, aur bar ka hatched right hissa dikhta hai. Chauthe step mein dono hisse bar ko bilkul 0 se 1 tak bhar dete hain, jo dikhata hai ki five-twelfths plus seven-twelfths barabar 1.

Quick check

3 mein se 1

Ek train ke time par aane ki probability 0.64 hai. Uske time par na aane ki probability kya hai?

Answer dekho

Answer 0.36

Time par aur time par nahi complementary hain:

1 - 0.64 = 0.36.

Impossible aur sure events

Do extreme cases picture poori karte hain. Fig. 14.4 wale spinner mein paanch barabar sectors hain jin par numbers 2, 4, 6, 8 aur 10 hain.

Iske odd number par rukne ki probability kya hai? Koi sector odd nahi hai, isliye koi favourable outcome nahi hai aur $P (odd) = \frac{0}{5} = 0.$ Jo event kisi experiment mein kabhi ho hi nahi sakta, woh impossible event (असंभव घटना) hai; uski probability 0 hai.

Ab even number ke baare mein poochho. Har sector even hai, isliye sabhi 5 outcomes favourable hain aur $P (even) = \frac{5}{5} = 1.$ Jo event har haal mein hona hi hai, outcome chahe jo ho, woh sure event hai, jise certain event bhi kehte hain (निश्चित घटना); uski probability 1 hai.

3/4Paanch mein se teen sectors fit hote hain: $P = \frac{3}{5} .$

Fig. 14.4Jaise-jaise zyada sectors event mein fit hote hain, uski probability 0 se 1 ki taraf chadhti hai.NCERT p. 207

Left mein ek spinner hai jiske top par pointer hai aur paanch barabar sectors par 2, 4, 6, 8 aur 10 likha hai. Right mein ek scale 0 se, jis par impossible likha hai, 1 tak jaata hai, jis par sure likha hai. Pehle step mein panel batata hai: paanch barabar sectors, sab par even labels. Doosre step mein event odd number hai: koi sector tick nahi hota, probability 0 over 5 yaani 0 hai, aur marker 0 par baitha hai. Teesre step mein event 5 se bada number hai: sectors 6, 8 aur 10 shade aur tick hote hain, probability three-fifths hai, aur marker 0.6 par jaata hai. Chauthe step mein event even number hai: har sector shade aur tick hota hai, probability 5 over 5 yaani 1 hai, aur marker 1 tak pahunchta hai.

Baaki har event in do extremes ke beech aata hai. Kisi bhi event ke liye favourable outcomes ki sankhya $m$ kam se kam 0 hoti hai aur zyada se zyada $n .$ $0 \leq m \leq n$ ko $n$ se divide karne par range rule milta hai.

Probability ki range

0 \leq P (E) \leq 1

(14.4)

Compare

Do extremes

Impossible event

koi favourable outcome nahi
$P = 0$
die par 0 aana

Sure event

har outcome favourable
$P = 1$
die par 1 se 6 tak ka number aana

Playing cards ka deck

Rummy aur teen patti jaise card games ek standard deck use karte hain, aur board questions bhi aksar yahi use karte hain. Ginne se pehle tumhe pata hona chahiye ki deck mein exactly kya-kya hai.

Deck chaar suits mein bata hota hai: clubs (♣), diamonds (♦), hearts (♥) aur spades (♠). Har suit mein 13 cards hote hain, isliye deck mein kul cards hain $4 \times 13 = 52.$
Clubs aur spades black mein chhape hote hain, diamonds aur hearts red mein: har colour ke 26 cards.
Har suit mein ek ace (A), number cards 2 se 10, ek jack (J), ek queen (Q) aur ek king (K) hota hai. Isliye har rank ke 4 cards hote hain.
Jacks, queens aur kings face cards hain: har suit mein 3, kul 12.

Jab deck well shuffled ho, to 52 cards mein se har ek ke nikalne ka chance same hota hai. Fig. 14.5 deck ko ek grid mein dikhata hai, har suit ki ek row, taaki tum kisi event ke cards ek nazar mein gin sako.

2/4Har suit mein J, Q aur K: $\frac{12}{52} = \frac{3}{13} .$

Fig. 14.5Chamakte cards gino aur 52 se divide karo.NCERT p. 207

Deck ke 52 cards chaar rows mein bichhe hain, har suit ki ek: spades, hearts, diamonds aur clubs, apne symbols ke saath, aur hearts aur diamonds red mein chhape hain. Har row ace, 2 se 10, jack, queen, king ke order mein chalti hai. Pehle step mein saare cards dikhte hain aur neeche likha hai 4 suits times 13 cards barabar 52 cards. Doosre step mein sirf jack, queen aur king ke columns chamakte rehte hain: 12 face cards, probability 12 over 52, yaani three-thirteenths. Teesre step mein sirf ace ka column chamakta hai: 4 aces, probability one-thirteenth. Chauthe step mein sirf hearts ki row chamakti hai: 13 hearts, probability one-quarter.

Example 14.4NCERT Example 4 · p. 207

Heart, ya heart nahi

Raghav 52 cards ka deck achhi tarah shuffle karta hai aur sabse upar wala card palat deta hai. Probability nikaalo ki yeh card (a) heart hai, (b) heart nahi hai. Part (b) do tareekon se karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

SimulationCard-drawing lab

52 cards ka ek virtual, well-shuffled deck. Har press 1, 10, 100 ya 1000 cards nikaalta hai, aur agle draw se pehle har card deck mein wapas chala jaata hai. Bars dikhate hain ki har suit kitni baar aaya, aur ek counter compare karta hai ki face card (jack, queen ya king) kitni baar aaya, uski theoretical probability three-thirteenths, lagbhag 0.23, se.

Fig. 14.6Shuffled deck se baar-baar card nikaalo, aur dekhte raho ki face card kitni baar aata hai.

Try karo

Pehle predict karo: 130 draws mein lagbhag kitne face cards expect karoge? Phir 10-10 karke 13 baar draw karo, aur compare karo.
Kuch baar ek saath 1000 draw karo. Kya face-card share $\frac{3}{13} \approx 0.23$ ke aas-paas tik jaata hai?

Quick check

3 mein se 1

Tum ek poore, shuffled deck se at random ek card uthate ho. Uske 7 hone ki probability kya hai?

\frac{1}{52}

\frac{1}{13}

\frac{7}{52}

\frac{4}{13}

Answer dekho

Answer (B) $\frac{1}{13}$

Har suit mein ek 7 hai, isliye 4 sevens:

\frac{4}{52} = \frac{1}{13} .

Value

\frac{1}{52}

kisi ek khaas 7 ke liye hai, jaise 7 of clubs.

Complement ka use

Complement rule aksar answer tak pahunchne ka sabse tez raasta hota hai, aur hamesha ek kaam ka check bhi. Agle chaar examples mein ise dhyaan se dekho.

Example 14.5NCERT Example 5 · p. 208

Chess final

Arjun aur Meera ek chess final khelte hain jismein draw allowed nahi hai: agar game draw hota hai, to tie-break games tab tak chalte hain jab tak koi jeet na jaaye. Arjun ke final jeetne ki probability 0.47 hai. Meera ke jeetne ki probability kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14.6NCERT Example 6 · p. 208

Do dost numbers chunte hain

Kavya aur Imran dono chupchaap 1 se 20 tak ka ek whole number chunte hain, har number equally likely hai, aur kisi ko doosre ki choice nahi pata. Probability nikaalo ki woh (a) alag-alag numbers chunte hain, (b) same number chunte hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14.7NCERT Example 7 · p. 209

Pehla paudha kaun lagayega?

School eco-club mein 36 members hain: 21 Class 9 se aur 15 Class 10 se. Van Mahotsav par pehla paudha kaun lagayega, yeh tay karne ke liye secretary 36 ek jaisi name slips ek steel dabbe mein daalta hai, use zor se hilata hai aur ek slip nikaalta hai. Probability nikaalo ki naam (a) Class 10 ke student ka hai, (b) Class 9 ke student ka hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14.8NCERT Example 8 · p. 209

Modaks ka ek box

Ganesh Chaturthi ke liye ek box mein same size aur shape ke 15 modaks hain: 6 chocolate, 5 coconut aur 4 kesar. Ek modak at random uthaya jaata hai. Probability nikaalo ki woh (a) coconut, (b) chocolate, (c) kesar hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek lucky-draw ticket ke kuch na jeetne ki probability 0.92 hai. Uske kuch jeetne ki probability kya hai?

Answer dekho

Answer 0.08

Kuch jeetna event kuch na jeetna ka complement hai:

1 - 0.92 = 0.08.

Do coins uchhaalna

Jab do coins ek saath uchhaale jaate hain, to unhe alag pehchaan do, maan lo ek ₹5 ka coin aur ek ₹10 ka coin, aur har outcome ko ordered pair ki tarah likho: pehla letter ₹5 coin ka aur doosra ₹10 coin ka. Tab (H, T) aur (T, H) alag outcomes hain. Ek mein ₹5 coin heads dikhata hai; doosre mein ₹10 coin.

Isliye chaar equally likely outcomes hain: (H, H), (H, T), (T, H) aur (T, T). Fig. 14.7 inhe ek table mein dikhata hai.

3/3Sirf (H, H) mein koi tail nahi: $P = 1 - \frac{1}{4} = \frac{3}{4} .$

Fig. 14.7Ordered pairs (H, T) aur (T, H) ko alag rakhte hain.NCERT p. 210

Ek table ke top par ₹10 coin ke faces, H aur T, hain aur side mein ₹5 coin ke faces, H aur T, jisse chaar cells bante hain: (H, H), (H, T), (T, H) aur (T, T). Pehle step mein panel kehta hai ki 4 equally likely outcomes hain aur (H, T) aur (T, H) same nahi hain. Doosre step mein cells (H, H) aur (T, T) shade hote hain, us event ke liye jismein dono coins same face dikhate hain: 4 mein se 2 cells, probability one-half. Teesre step mein at least ek tail wale teen cells shade hote hain aur (H, H) ko kaata jaata hai kyunki sirf usmein koi tail nahi hai, isliye probability 1 minus one-quarter yaani three-quarters hai.

Example 14.9NCERT Example 9 · p. 210

Do coins, do events

Ritu ek ₹5 coin aur ek ₹10 coin ek saath uchhaalti hai. Probability nikaalo ki (a) dono coins same face dikhate hain, (b) use at least ek tail milti hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Do coins ek saath uchhaale jaate hain. Kaunsa statement sahi hai?

Do heads, do tails aur ek-ek equally likely hainDo tails ka chance do heads se zyada haiAt least ek head ki probability

\frac{1}{2}

haiEk head aur ek tail ka chance do heads se dugna hai

Answer dekho

Answer (D) Ek head aur ek tail ka chance do heads se dugna hai

Ek-ek mein (H, T) aur (T, H) aate hain, isliye uski probability hai

\frac{2}{4},

jo (H, H) ke

\frac{1}{4}

ka dugna hai. Teeno tarah ke results equally likely nahi hain, aur at least ek head hai

\frac{3}{4} .

Ginti se aage: lengths aur areas

Ab tak har example mein experiment ke outcomes ki sankhya finite thi jise hum gin sakte the. Kuch experiments alag hote hain. Tum bus stop par jis pal pahunchte ho, woh ek interval ka koi bhi instant ho sakta hai, aur baarish ki boond khet ke kisi bhi point par gir sakti hai. Aise experiments ke infinitely many outcomes hote hain, isliye hum favourable outcomes gin nahi sakte. Iski jagah hum naapte hain: agar har point equally likely hai, to probability hai favourable length (ya area) divided by total length (ya area).

4/4Favourable area hai $120 m^{2} :$ $P = \frac{120}{1500} = \frac{2}{25} .$

Fig. 14.8Infinitely many outcomes hon to counts ki jagah lengths ya areas compare karo.NCERT Fig. 14.2 · p. 211

Upar ek number line 0 se 15 minutes tak arrival times 3-3 ke steps mein mark karti hai. Pehle step mein sirf line dikhti hai. Doosre step mein 9 se 15 minutes wala hissa green ho jaata hai aur us par wait at most 6 minutes likha hota hai, probability 6 over 15 yaani two-fifths. Teesre step mein neeche 50 metres by 30 metres ka rectangular field aata hai, jahan seed packets gire hain wahan kuch crosses hain, aur uske andar 12 metres by 10 metres ka pond hai; panel field ka area 1500 square metres batata hai. Chauthe step mein pond ki outline moti ho jaati hai aur panel uska area, 120 square metres, aur probability 120 over 1500 yaani two twenty-fifths jodta hai.

Example 14.10NCERT Example 10 · p. 210

Ferry ka intezaar

Ghat se har 15 minute mein ek ferry chhootti hai. Tum timetable dekhe bina kisi random pal ghat pahunchte ho. Tumhare zyada se zyada 6 minute wait karne ki probability kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14.11NCERT Example 11 · p. 211

Seed packet aur pond

Ek drone seed packet giraata hai jo 50 m lambe aur 30 m chaude rectangular field mein kahin bhi equally likely gir sakta hai. Field ke andar 12 m by 10 m ka rectangular pond hai. Packet ke pond mein girne ki probability kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

School ki bell 10:00 aur 10:20 ke beech kisi random pal bajti hai. Uske 10:15 aur 10:20 ke beech bajne ki probability kya hai?

Answer dekho

Answer 0.25

20 mein se favourable length 5 minutes:

\frac{5}{20} = \frac{1}{4} .

Do buyers, do dice

Ek hi random draw ko alag-alag rules se judge kiya ja sakta hai. Agle example mein aamon ke ek crate ko do buyers check karte hain jinhe alag cheezein chahiye; Fig. 14.9 dikhata hai ki har ek kya accept karta hai.

3/3Rotten ke alaawa sab: $\frac{188}{200} = 0.94.$

Fig. 14.9Ek hi random aam, do alag rules se judge kiya gaya.NCERT p. 212

Ek lamba bar crate ke 200 aamon ko dikhata hai. Uska plain left hissa 160 perfect aam hai, striped beech ka hissa 28 bruised aam hai aur cross-hatched right end 12 rotten aam hai, neeche ek key ke saath. Pehle step mein sirf bar dikhta hai. Doosre step mein perfect hisse ke upar ek bracket par fruit-seller 160 likha hai, probability 160 over 200 yaani 0.8. Teesre step mein perfect aur bruised hisson ke upar ek lamba bracket aata hai jis par juice stall 188 likha hai, probability 188 over 200 yaani 0.94.

Example 14.12NCERT Example 12 · p. 212

Do buyers ke liye aam

Ek crate mein 200 aam hain: 160 perfect hain, 28 thode bruised hain aur 12 rotten hain. Fruit-seller sirf perfect aam lega. Juice stall koi bhi aam lega jo rotten na ho. Crate se ek aam at random uthaya jaata hai. Probability nikaalo ki woh (a) fruit-seller ko, (b) juice stall ko acceptable hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Do dice se outcomes ka bada set milta hai. Alag colours ke dice lo, maan lo red aur green, aur har outcome ko ordered pair (red, green) ki tarah likho. Red die 6 mein se koi bhi number dikha sakta hai, aur har ek ke liye green die bhi 6 mein se koi bhi, isliye $6 \times 6 = 36$ equally likely outcomes hain. Pair (2, 5) ka matlab hai red 2 aur green 5; yeh (5, 2) se alag outcome hai.

4/47 ke paas wale sums ke cells zyada hain, isliye unka chance zyada hai.

Fig. 14.1036 cells mein se har ek equally likely hai, lekin sums nahi: 7 chhe tareekon se aa sakta hai, 2 aur 12 sirf ek se.NCERT Fig. 14.3 · p. 213

Ek grid ke top par green die ke numbers 1 se 6 hain aur side mein red die ke numbers 1 se 6; 36 cells mein se har ek apne do numbers ka sum dikhata hai. Pehle step mein panel kehta hai ki 36 outcomes hain aur har cell sum dikhata hai. Doosre step mein sum 9 wale chaar cells shade aur ring hote hain, jo diagonally chalte hain, aur panel probability 4 over 36 yaani one-ninth batata hai. Teesre step mein main diagonal ke chhe doubles shade aur ring hote hain, probability 6 over 36 yaani one-sixth. Chauthe step mein grid ke paas ek bar chart dikhata hai ki har sum kitne cells se aata hai: sums 2 se 12 ke liye 1, 2, 3, 4, 5, 6, 5, 4, 3, 2 aur 1, aur 7 ka bar sabse ooncha hai.

Example 14.13NCERT Example 13 · p. 212

Do dice par sums

Ek red die aur ek green die ek saath roll kiye jaate hain. Chance nikaalo ki dono numbers ka sum (a) 9, (b) 1, (c) 1 se zyada hai. Saath hi (d) double ka chance bhi nikaalo, yaani dono dice par same number.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

SimulationTwo-dice lab

Do virtual dice jo ek press mein ek saath 1, 10, 100 ya 1000 baar roll hote hain. Ek bar chart dikhata hai ki 2 se 12 tak har sum kitni baar aaya, theoretical values ke saath, aur ek counter compare karta hai ki total 9 kitni baar aaya, uski theoretical probability one-ninth se.

Fig. 14.11Do dice ko batches mein roll karo aur compare karo ki har sum kitni baar aata hai, theory ke saath.

Try karo

Ek baar mein 10 roll karo, teen baar. 9s ka share $\frac{1}{9} \approx 0.11$ se kitna door hai?
Ab ek baar mein 1000 roll karo. Kaunsa sum sabse common hai, aur kya bars ki shape upar wale figure ke bar chart se match karti hai?

Quick check

3 mein se 1

Do dice roll kiye jaate hain. Sum 5 hone ki probability kya hai?

Answer dekho

Answer 0.1111111111111111

(1, 4), (2, 3), (3, 2) aur (4, 1):

\frac{4}{36} = \frac{1}{9} \approx 0.111.

Practice · 25 questions

Practice 14.1: answers ke saath

1
Quick answerEasyNCERT Exercise 14.1, Q1
Blanks bharo.
1. Event $E,$ chahe jo bhi ho, sum $P (E) + P (\overset{ˉ}{E})$ hamesha hota hai
  Answer dekho
  Answer 1
  Event aur uska complement sabhi outcomes aapas mein baant lete hain.
2. Jo event kabhi ho hi nahi sakta, uski probability hoti hai aur use event kehte hain
  Answer dekho
  Answer (1) 0, (2) impossible
  Koi outcome favourable nahi hai, isliye $m = 0.$
3. Jo event hona pakka hai, uski probability hoti hai aur use event kehte hain
  Answer dekho
  Answer (1) 1, (2) sure
  Har outcome favourable hai, isliye $m = n .$
4. Agar tum kisi experiment ke har elementary event ki probabilities jodo, to milta hai . Koi bhi probability se kam ya se zyada nahi hoti
  Answer dekho
  Answer (1) 1, (2) 0, (3) 1
  Elementary events milkar har outcome ko ek baar cover karte hain, aur $0 \leq m \leq n$ se milta hai $0 \leq P (E) \leq 1.$
2
Short answerEasyNCERT Exercise 14.1, Q2
Har experiment mein tay karo ki dono outcomes equally likely hain ya nahi, aur reason do.
1. Achhi tarah mile ek bowl se ek slip nikaali jaati hai, jismein do ek jaisi slips hain, ek par Yes likha hai aur ek par No. Result Yes ya No hai.
  equally likelyequally likely nahi
  Solution dikhao
  Equally likely: slips ek jaisi hain aur achhi tarah mili hain, isliye har ek ke nikalne ka chance same hai.
2. Ek kabaddi raider doosre half mein jaati hai. Woh point score karti hai ya pakdi jaati hai.
  equally likelyequally likely nahi
  Solution dikhao
  Equally likely nahi: result uski skill aur defenders par depend karta hai, isliye barabar chances ki koi wajah nahi hai.
3. July ke kisi din Chennai mein baarish hoti hai ya nahi hoti.
  equally likelyequally likely nahi
  Solution dikhao
  Equally likely nahi: mausam season aur us din ki conditions par depend karta hai; koi cheez dono results ko balance nahi karti.
4. Ek ordinary die roll kiya jaata hai. Woh even number dikhata hai ya odd number.
  equally likelyequally likely nahi
  Solution dikhao
  Equally likely: har result chhe equally likely faces mein se teen cover karta hai (2, 4, 6 aur 1, 3, 5), isliye har ek ki probability hai $\frac{1}{2} .$
Short answerEasyNCERT Exercise 14.1, Q3
Kabaddi match se pehle referee coin uchhaalta hai aur toss jeetne wala captain chunta hai ki kaunsi team pehle raid karegi. Yeh decide karne ka fair tareeka kyun hai?
Solution dikhao
At random uchhaala gaya fair coin heads ya tails par equally likely girta hai. Captain chahe jo face call kare, uske toss jeetne ka chance hai $\frac{1}{2},$ bilkul doosre captain jitna. Kisi team ka favour nahi hota, aur koi result ko control nahi kar sakta.
Quick answerEasyNCERT Exercise 14.1, Q4
Inmein se kaun kisi event ki probability nahi ho sakti?
$\frac{7}{9}$ $108 %$ $0.001$ $0$
Answer dekho
Answer (B) $108 %$
Har probability 0 se 1 ke beech hoti hai. $108 % = 1.08,$ jo 1 se zyada hai. Value 0 allowed hai: yeh impossible event ki probability hai.
NumericalEasyNCERT Exercise 14.1, Q5
Agar $P (E) = 0.17,$ to not $E$ ki probability kya hai?
Solution dikhao
$P (\overset{ˉ}{E}) = 1 - 0.17 = 0.83.$
6
Short answerEasyNCERT Exercise 14.1, Q6
Dadi ke dabbe ka har laddoo coconut laddoo hai. Arjun aankhein band karke ek uthata hai. Chance nikaalo ki usne uthaya hai
1. besan laddoo;
  Solution dikhao
  Dabbe mein koi laddoo besan ka nahi hai, isliye yeh impossible event hai: $P = 0.$
2. coconut laddoo.
  Solution dikhao
  Har laddoo coconut ka hai, isliye yeh sure event hai: $P = 1.$
NumericalEasyNCERT Exercise 14.1, Q7
At random chune gaye do logon ke week ke alag-alag dinon par paida hone ki probability hai $\frac{6}{7} .$ Unke week ke same din paida hone ki probability kya hai?
Solution dikhao
$P (same day) = 1 - \frac{6}{7} = \frac{1}{7} \approx 0.143.$
8
Short answerEasyNCERT Exercise 14.1, Q8
Riya ke pouch mein same length ki 7 green aur 4 yellow pencils hain. Woh bina dekhe ek nikaalti hai. Kitna likely hai ki woh
1. green ho;
  Solution dikhao
  $n = 7 + 4 = 11$ aur $m = 7,$ isliye $P (green) = \frac{7}{11} .$
2. green na ho.
  Solution dikhao
  $P (not green) = 1 - \frac{7}{11} = \frac{4}{11},$ yaani yellow pencils ka share.
9
Short answerEasyNCERT Exercise 14.1, Q9
Birthday party mein ek jar mein ek jaise wrappers wali 20 toffees hain: 6 mango, 9 orange aur 5 lime. Pehla guest bina dekhe ek uthata hai. Probability kya hai ki woh
1. mango ho;
  Solution dikhao
  $n = 20,$ isliye $P (mango) = \frac{6}{20} = \frac{3}{10} .$
2. orange ho;
  Solution dikhao
  $P (orange) = \frac{9}{20} .$
3. lime na ho.
  Solution dikhao
  $20 - 5 = 15$ toffees lime nahi hain, isliye $P = \frac{15}{20} = \frac{3}{4} .$
10
Short answerMediumNCERT Exercise 14.1, Q10
Meena ki mitti ki gullak mein 150 coins hain: ₹1 ke 60, ₹2 ke 45, ₹5 ke 30 aur ₹10 ke 15. Woh use apne bed ke upar tab tak hilaati hai jab tak slot se ek coin bahar na gir jaaye; maan lo har coin ke girne ka chance same hai. Probability kya hai ki coin
1. ₹2 ka ho;
  Solution dikhao
  $n = 150,$ aur 45 coins ₹2 ke hain, isliye probability hai $\frac{45}{150} = \frac{3}{10} .$
2. ₹10 ke coin ke alaawa kuch bhi ho?
  Solution dikhao
  $P = 1 - \frac{15}{150} = \frac{135}{150} = \frac{9}{10} .$
NumericalEasyNCERT Exercise 14.1, Q11
Ek nursery tray mein 7 rose saplings aur 11 marigold saplings hain jo phool aane se pehle ek jaise dikhte hain. Ek maali at random ek uthata hai. Uske rose sapling hone ki probability kya hai?
Solution dikhao
$n = 7 + 11 = 18$ aur $m = 7,$ isliye $P (rose) = \frac{7}{18} \approx 0.39.$
12
Short answerMediumNCERT Exercise 14.1, Q12
Ek wheel mein 1 se 10 number wale 10 barabar sectors hain, aur uska arrow kisi bhi sector par rukne ke liye equally likely hai. Probability nikaalo ki woh rukta hai
1. 7;
  Solution dikhao
  Ek sector par 7 hai: $P = \frac{1}{10} .$
2. prime number par;
  Solution dikhao
  Primes: 2, 3, 5, 7, isliye $P = \frac{4}{10} = \frac{2}{5} .$
3. aise number par jo kam se kam 4 ho;
  Solution dikhao
  4 se 10 tak 7 numbers hain: $P = \frac{7}{10} .$
4. 11 se chhote number par.
  Solution dikhao
  Saare 10 sectors: $P = 1,$ yaani sure event.
13
Short answerEasyNCERT Exercise 14.1, Q13
Ek die ek baar roll kiya jaata hai. Probability nikaalo ki aata hai
1. 6 ka factor;
  Solution dikhao
  Die par 6 ke factors: 1, 2, 3, 6, isliye $P = \frac{4}{6} = \frac{2}{3} .$
2. 1 aur 5 ke bilkul beech ka number (dono ko chhod kar);
  Solution dikhao
  2, 3 aur 4, isliye $P = \frac{3}{6} = \frac{1}{2} .$
3. perfect square.
  Solution dikhao
  1 aur 4, isliye $P = \frac{2}{6} = \frac{1}{3} .$
14
Short answerMediumNCERT Exercise 14.1, Q14
52 cards ke achhi tarah shuffle kiye pack se ek card at random nikaala jaata hai. Probability kya hai ki woh
1. black queen ho;
  Solution dikhao
  Spades aur clubs ki queens: $\frac{2}{52} = \frac{1}{26} .$
2. face card na ho;
  Solution dikhao
  $52 - 12 = 40$ cards: $\frac{40}{52} = \frac{10}{13} .$
3. black face card ho;
  Solution dikhao
  Spades ke aur clubs ke J, Q, K: $\frac{6}{52} = \frac{3}{26} .$
4. 7 of clubs ho.
  Solution dikhao
  Bilkul ek card: $\frac{1}{52} .$
15
Short answerMediumNCERT Exercise 14.1, Q15
Kabir clubs ke ace se 6 tak ke chhe cards leta hai, unhe mix karta hai aur ek row mein ulta (face down) rakh deta hai. Woh at random ek card palatta hai.
1. Uske 4 hone ki probability kya hai?
  Solution dikhao
  6 equally likely cards, unmein se ek 4 hai: $P = \frac{1}{6} .$
2. Woh 4 hi nikalta hai, aur use alag rakh diya jaata hai. Phir bache hue cards mein se ek at random palta jaata hai. Uske ace hone ka chance nikaalo.
  Solution dikhao
  5 cards bache hain aur unmein ek ace hai: $P = \frac{1}{5} .$
3. Part (b) mein, doosre card ke 4 hone ki probability kya hai?
  Solution dikhao
  4 ab cards mein hai hi nahi: yeh impossible event hai, $P = 0.$
NumericalEasyNCERT Exercise 14.1, Q16
300 phone chargers ke ek batch mein 15 kaam nahi karte; kharaab chargers baaki jaise hi dikhte hain. Ek charger at random liya jaata hai. Uske kaam karne ki probability kya hai?
Solution dikhao
$300 - 15 = 285$ chargers kaam karte hain, isliye $P = \frac{285}{300} = \frac{19}{20} = 0.95.$
17
Short answerMediumNCERT Exercise 14.1, Q17
Ek tokri mein 32 aam hain. Unmein se 8 andar se kharaab hain, halaanki bahar se theek dikhte hain.
1. Ek aam at random liya jaata hai. Uske kharaab hone ki probability kya hai?
  Solution dikhao
  $P = \frac{8}{32} = \frac{1}{4} .$
2. (a) mein liya gaya aam achha nikalta hai aur kha liya jaata hai. Phir tokri mein bache aamon mein se doosra aam at random chuna jaata hai. Uske achhe hone ka chance nikaalo.
  Solution dikhao
  31 aam bache hain, jinmein se $24 - 1 = 23$ achhe hain, isliye $P = \frac{23}{31} .$
18
Short answerMediumNCERT Exercise 14.1, Q18
Ek tambola evening mein caller ke bag mein 1 se 60 number wale tokens hain. Pehla token at random nikaala jaata hai. Probability nikaalo ki uska number
1. single-digit number ho;
  Solution dikhao
  1 se 9 tak 9 numbers hain: $P = \frac{9}{60} = \frac{3}{20} .$
2. perfect cube ho;
  Solution dikhao
  1, 8 aur 27 (agla cube, 64, bahut bada hai): $P = \frac{3}{60} = \frac{1}{20} .$
3. 6 ka multiple ho.
  Solution dikhao
  6, 12, …, 60 yaani 10 numbers: $P = \frac{10}{60} = \frac{1}{6} .$
19
Short answerEasyNCERT Exercise 14.1, Q19
Ek cube ke chhe faces par letters B, A, N, A, N aur A likhe hain. Cube ek baar roll kiya jaata hai. Probability nikaalo ki top face par aata hai
1. A;
  Solution dikhao
  Chhe mein se teen faces par A hai: $P = \frac{3}{6} = \frac{1}{2} .$
2. B.
  Solution dikhao
  Ek face par B hai: $P = \frac{1}{6} .$
NumericalHardNCERT Exercise 14.1, Q20
(Board exam ke liye nahi.) Ek rectangular garden 14 m lamba aur 10 m chauda hai, aur uske andar 3.5 m radius ka ek circular fountain hai. Hawa se uda ek beej garden ke kisi random point par girta hai. Uske fountain mein girne ki probability kya hai? Use karo $π = \frac{22}{7} .$
Solution dikhao
Garden: $14 \times 10 = 140 m^{2} .$ Fountain: $\frac{22}{7} \times {3.5}^{2} = 38.5 m^{2} .$
$P = \frac{38.5}{140} = \frac{11}{40} = 0.275.$
21
Short answerEasyNCERT Exercise 14.1, Q21
Ek shelf par 200 biscuit packets mein se 16 ki expiry date nikal chuki hai, jo flap ke andar chhapi hai. Meera ka rule simple hai: packet date ke andar ho to rakh leti hai, warna lauta deti hai. Uska bhai shelf se at random ek packet laata hai. Probability nikaalo ki Meera
1. use rakh leti hai;
  Solution dikhao
  $200 - 16 = 184$ packets date ke andar hain: $P = \frac{184}{200} = \frac{23}{25} .$
2. use lauta deti hai.
  Solution dikhao
  $P = 1 - \frac{23}{25} = \frac{2}{25} .$
22
Long answerHardNCERT Exercise 14.1, Q22
Do dice ek saath roll kiye jaate hain, aur hum bade aur chhote number ka difference note karte hain (barabar hon to 0).
1. Differences 0, 1, 2, 3, 4 aur 5 ki probabilities ka table banao. Difference 1 hone ki probability kya hai?
  Solution dikhao
  Har difference ke pairs ginne par:
  Table 14.1Do dice ka difference: ordered pairs ki sankhya aur probability $P$
  Difference Pairs $P$
  0 6 $\frac{1}{6}$
  1 10 $\frac{5}{18}$
  2 8 $\frac{2}{9}$
  3 6 $\frac{1}{6}$
  4 4 $\frac{1}{9}$
  5 2 $\frac{1}{18}$
  Counts ka total 36 hai. Difference 1 aata hai (1, 2), (2, 1), (2, 3), (3, 2), …, (5, 6), (6, 5) ke liye: 10 pairs, isliye $P = \frac{10}{36} = \frac{5}{18} .$
2. Ek classmate kehta hai: "Difference sirf 0, 1, 2, 3, 4 ya 5 ho sakta hai, isliye har ek ki probability hai $\frac{1}{6} .$ " Kya tum agree karte ho? Samjhao.
  Solution dikhao
  Nahi. Chhe differences equally likely nahi hain; equally likely sirf 36 ordered pairs hain. Table se, difference 1 ki probability $\frac{5}{18}$ hai lekin difference 5 ki sirf $\frac{1}{18} .$
23
Short answerMediumNCERT Exercise 14.1, Q23
Priya ek coin teen baar uchhaalti hai aur har baar result note karti hai. Agar use bilkul do heads milte hain to woh prize jeetti hai. Probability nikaalo ki woh
1. jeetti hai;
  Solution dikhao
  Bilkul do heads: HHT, HTH, THH, yaani 8 equally likely results mein se 3: $P = \frac{3}{8} .$
2. nahi jeetti.
  Solution dikhao
  $P = 1 - \frac{3}{8} = \frac{5}{8} .$
24
Short answerMediumNCERT Exercise 14.1, Q24
Ek die do baar roll kiya jaata hai. Probability nikaalo ki
1. dono rolls par even numbers aate hain;
  Solution dikhao
  Pehle roll ke liye 3 even faces aur doosre ke liye 3: $3 \times 3 = 9$ pairs, isliye $P = \frac{9}{36} = \frac{1}{4} .$
2. kam se kam ek roll par odd number aata hai.
  Solution dikhao
  Yeh (a) ka complement hai: $P = 1 - \frac{1}{4} = \frac{3}{4} .$
25
Short answerMediumNCERT Exercise 14.1, Q25
Batao ki har argument sahi hai ya nahi, aur reason do.
1. Jab die roll hota hai, to ya to 6 aata hai ya nahi aata. Isliye 6 ki probability hai $\frac{1}{2} .$
  sahisahi nahi
  Solution dikhao
  Sahi nahi. 6 aana aur 6 na aana equally likely nahi hain: 6 na aana chhe equally likely faces mein se paanch cover karta hai. Asal mein $P (6) = \frac{1}{6} .$
2. Well-shuffled deck se nikaala gaya card ya to red hota hai ya black. Isliye uske red hone ki probability hai $\frac{1}{2} .$
  sahisahi nahi
  Solution dikhao
  Sahi, lekin sirf isliye ki dono colours ke cards barabar sankhya mein hain: 52 mein se 26 red, isliye $P (red) = \frac{26}{52} = \frac{1}{2} .$

Difference	Pairs	$P$
0	6	$\frac{1}{6}$
1	10	$\frac{5}{18}$
2	8	$\frac{2}{9}$
3	6	$\frac{1}{6}$
4	4	$\frac{1}{9}$
5	2	$\frac{1}{18}$

14.2 Summary

4/4Unki probabilities ka total 1 hota hai.

Fig. 14.12Chaar tools is chapter ki lagbhag har problem solve kar dete hain.NCERT p. 217

Chaar tiles ek ke baad ek aati hain. Tile 1, Count, dikhati hai P of E barabar favourable outcomes over sabhi outcomes. Tile 2, Range, jodti hai 0 at most P of E, jo at most 1 hai, jismein 0 par impossible aur 1 par sure likha hai. Tile 3, Complement, jodti hai P of E plus P of not E barabar 1. Tile 4, Elementary events, jodti hai ki sabhi elementary events ki probabilities ka total 1 hota hai.

Recap

Summary aur key points

Outcomes equally likely tab hote hain jab set-up mein koi cheez kisi ka favour nahi karti; outcomes aise list karo ki yeh sach rahe (Equally likely outcomes).
Theoretical probability: $P (E) = \frac{m}{n},$ yaani favourable outcomes over sabhi outcomes, equally likely outcomes ke liye (Theoretical probability).
Elementary event mein bilkul ek outcome hota hai, aur sabhi elementary events ki probabilities ka total 1 hota hai (Elementary events).
Kisi bhi event ke liye $P (E) + P (\overset{ˉ}{E}) = 1,$ isliye $P (\overset{ˉ}{E}) = 1 - P (E)$ (Complementary events).
Impossible event ki probability 0 hoti hai aur sure event ki 1; hamesha $0 \leq P (E) \leq 1$ (Impossible aur sure events).
Deck mein 52 cards hote hain: 13-13 ke 4 suits, 26 red aur 26 black, har rank ke 4 aur 12 face cards (Playing cards ka deck).
Do coins (4 outcomes) aur do dice (36 outcomes) ke liye ordered pairs use karo; do dice ke sums equally likely nahi hote (Do coins uchhaalna, Do buyers, do dice).
Board syllabus se aage (Ginti se aage: lengths aur areas): jab outcomes ek length ya area bharte hain, to favourable length ya area ko total se divide karo.
Bahut saare trials hon to experimental probability theoretical probability ke kareeb aane lagti hai (Key idea).

Flash cards · 7

Key terms

1 / 7

Board exam ki practice

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 14.1, Q1	Practice Q1
NCERT Exercise 14.1, Q2	Practice Q2
NCERT Exercise 14.1, Q3	Practice Q3
NCERT Exercise 14.1, Q4	Practice Q4
NCERT Exercise 14.1, Q5	Practice Q5
NCERT Exercise 14.1, Q6	Practice Q6
NCERT Exercise 14.1, Q7	Practice Q7
NCERT Exercise 14.1, Q8	Practice Q8
NCERT Exercise 14.1, Q9	Practice Q9
NCERT Exercise 14.1, Q10	Practice Q10
NCERT Exercise 14.1, Q11	Practice Q11
NCERT Exercise 14.1, Q12	Practice Q12
NCERT Exercise 14.1, Q13	Practice Q13
NCERT Exercise 14.1, Q14	Practice Q14
NCERT Exercise 14.1, Q15	Practice Q15
NCERT Exercise 14.1, Q16	Practice Q16
NCERT Exercise 14.1, Q17	Practice Q17
NCERT Exercise 14.1, Q18	Practice Q18
NCERT Exercise 14.1, Q19	Practice Q19
NCERT Exercise 14.1, Q20	Practice Q20
NCERT Exercise 14.1, Q21	Practice Q21
NCERT Exercise 14.1, Q22	Practice Q22
NCERT Exercise 14.1, Q23	Practice Q23
NCERT Exercise 14.1, Q24	Practice Q24
NCERT Exercise 14.1, Q25	Practice Q25

Quick check: MCQs

Theoretical probability3 questions
Elementary events3 questions
Complementary events3 questions
Impossible aur sure events3 questions
Playing cards ka deck3 questions
Complement ka use3 questions
Do coins uchhaalna2 questions
Ginti se aage: lengths aur areas2 questions
Do buyers, do dice3 questions

Glossary · 11

Key terms (English + Hindi)

at randomयादृच्छयाyadrichchhaya: Bina kisi bias ya control ke chuna ya phenka gaya, taaki har outcome ka natural chance bana rahe.Chapter mein pehli baar yahan aaya →

complementary eventsपूरक घटनाएँpoorak ghatnayen: Event $E$ aur event not $E,$ jiska symbol hai $\overset{ˉ}{E};$ inmein se bilkul ek hota hai, isliye $P (E) + P (\overset{ˉ}{E}) = 1.$

elementary eventप्रारंभिक घटनाprarambhik ghatna: Aisa event jo experiment ke bilkul ek outcome se bana ho.
equally likelyसमप्रायिकsamprayik: Hone ka same chance hona, kyunki set-up mein koi cheez ek outcome ka doosre se zyada favour nahi karti.Chapter mein pehli baar yahan aaya →
eventघटनाghatna: Kisi experiment ke outcomes ka collection, jaise die par even number.Chapter mein pehli baar yahan aaya →
experimental probabilityप्रायोगिक प्रायिकताprayogik prayikta: Trials ka woh fraction jismein event sach mein hua.Chapter mein pehli baar yahan aaya →

face card: Jack, queen ya king; 52 cards ke deck mein aise 12 hote hain.Chapter mein pehli baar yahan aaya →
favourable outcomeअनुकूल परिणामanukool parinam: Aisa outcome jo diye gaye event ko hone deta hai.Chapter mein pehli baar yahan aaya →

impossible eventअसंभव घटनाasambhav ghatna: Aisa event jiska koi favourable outcome nahi; uski probability 0 hai.Chapter mein pehli baar yahan aaya →

sure eventनिश्चित घटनाnishchit ghatna: Aisa event jiske liye har outcome favourable hai; uski probability 1 hai.Chapter mein pehli baar yahan aaya →

theoretical probabilityसैद्धांतिक प्रायिकताsaiddhantik prayikta: Equally likely outcomes ke liye $P (E) = \frac{m}{n} :$ favourable outcomes divided by sabhi possible outcomes.

Probability: Class 10 Maths Chapter 14 Hinglish mein

Is chapter mein aap kya seekhenge

14.1 Probability — Ek Theoretical Approach

Equally likely outcomes

Theoretical probability

Elementary events

Complementary events

Impossible aur sure events

Playing cards ka deck

Complement ka use

Do coins uchhaalna

Ginti se aage: lengths aur areas

Do buyers, do dice

Practice 14.1: answers ke saath

14.2 Summary

Summary aur key points

Board exam ki practice

Practice questions answers ke saath (NCERT “Board exam ki practice”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye