Probability: Class 11 Maths Chapter 14 Hinglish mein

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

Published 22 Oct 2026 · Updated 22 Oct 2026

75 min
2 simulations
43 practice questions
20 MCQs

Is chapter mein aap kya seekhenge

Events ko sample space ke subsets ki tarah describe karna, aur kisi outcome se decide karna ki event hua ya nahi
Events ko impossible, sure, simple ya compound mein classify karna, aur set operations se 'not A', 'A or B', 'A and B' aur 'A but not B' banana
Check karna ki diye gaye events mutually exclusive hain, exhaustive hain, ya dono
Probability ke teen axioms batana, check karna ki outcomes ko di gayi probabilities valid hain ya nahi, aur outcome probabilities jodkar kisi event ki probability nikalna
Equally likely outcomes ke liye P(E) = n(E)/n(S) derive karna, aur ise lagane ke liye counting, permutations aur combinations use karna
'A or B' ke liye addition rule aur 'not A' ke liye complement rule prove karna aur lagana, mutually exclusive events wale special case ke saath
Coins, dice, cards, committees aur random orders par board- aur entrance-level problems solve karna, teen events wale addition rule ke saath

NCERT Mathematics Class 11, Chapter 14, pp. 289–313 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

14.1 Event

Cricket match se pehle dono captains toss ke liye bahar aate hain, aur tab se commentators chances ki baat karte hain: baarish ke, bade total ke, collapse ke. Yeh chapter aisi baaton ko do steps mein exact banata hai. Pehle hum har us 'ghatna' ko, jiski humein parwaah ho sakti hai, outcomes ke ek set ki tarah describe karte hain, taaki Chapter 1 ka sets ka algebra saara hisaab-kitaab sambhal le. Phir hum teen chhote rules batate hain, axioms, jinhe har probability ko follow karna hi padta hai, aur unhi se woh har formula derive karte hain jo tum use karoge.

Coin aur die ke saath hi raho. Maan lo tumhe sirf itni parwaah hai ki head ke saath even number aata hai ya nahi. Teen outcomes yeh dete hain: H2, H4 aur H6. Inhe ek set mein ikattha karo, $E = {H 2, H 4, H 6},$ jo $S$ ka ek subset hai. Result ke baare mein koi bhi haan-ya-na wala sawaal isi tarah ek subset chun leta hai, jaisa Fig. 14.1 dikhata hai, aur har subset kisi na kisi aise sawaal ka jawaab hai.

2/5'Head aur even number' hai $E = {H 2, H 4, H 6} .$

Fig. 14.1Event bas sample space ka ek subset hai.NCERT p. 289

Do rows mein baarah tiles coin toss karne aur die roll karne ke outcomes dikhati hain: upar H1 se H6, neeche T1 se T6. Pehle step mein koi tile light nahi hai aur label kehta hai ki S mein 12 outcomes hain. Doosre step mein head ke saath even number wale event ke liye H2, H4 aur H6 light hote hain. Teesre step mein 4 se bade number ke liye H5, H6, T5 aur T6 light hote hain. Chauthe step mein kuch light nahi hota, kyunki kisi outcome mein 7 nahi aata, isliye event empty set hai. Aakhri step mein result H4 ko outlined event E ke andar tick se mark kiya gaya hai, yaani E ho gaya hai.

Table 14.1Coin-aur-die experiment ke kuch events

Event shabdon mein	$S$ ka uska subset
6 ke saath tail	${T 6}$
odd number, koi bhi face	${H 1, H 3, H 5, T 1, T 3, T 5}$
3 se chhote number ke saath head	${H 1, H 2}$
7 se chhota number	$S$ (saare 12 outcomes)
6 se bade number ke saath tail	$ϕ$ (koi outcome nahi)

Definition

Event

Event (घटना) kisi experiment ke outcomes ka ek set hai, yaani uske sample space $S$ ka ek subset. Har subset event hai, $ϕ$ aur khud $S$ bhi.

ExampleDie ke saath toss kiye coin ke liye, ${H 2, H 4, H 6}$ event 'head aur even number' hai.

Kyunki har subset event hai, $n$ outcomes wale sample space mein $2^{n}$ events hote hain, uske power set (Chapter 1) ke har subset ke liye ek. Chhote se coin-aur-die experiment mein bhi $2^{12} = 4096$ events hain, isiliye hum events ko sab list karne ke bajaye rules se describe karte hain.

14.1.1 Event ka hona (occurrence)

Experiment ek baar karo. Maan lo coin par head aata hai aur die par 4, to outcome H4 hai. Yeh outcome $E = {H 2, H 4, H 6},$ ka element hai, aur hum kehte hain ki $E$ ho gaya hai (Fig. 14.1 ka aakhri step). Agar result T4 hota, to outcome $E$ ke bahar hota aur $E$ nahi hota.

Yaani event ka hona bas membership hai. Ek trial ek outcome $ω$ par khatam hota hai; har woh event jismein $ω$ hai, us trial mein ho gaya, aur jo event $ω$ ko chhod deta hai, woh nahi hua. Ek outcome ek saath kai events karwa deta hai: H4 se 'head', 'even number' aur '3 se bada number' bhi ho jaate hain, kyunki H4 teeno subsets mein hai.

14.1.2 Events ke types

Events ko is hisaab se baanta jaata hai ki unmein kitne sample points hain. Ek spinner lo jismein 1 se 8 tak numbered aath barabar sectors hain, to $S = {1, 2, \dots, 8},$ aur chaar events dekho (Fig. 14.2).

Impossible event. 'Pointer 8 se bade number par rukta hai' mein $S$ ka koi outcome nahi hai, isliye set ke roop mein yeh $ϕ$ hai. Empty set ko impossible event kehte hain: koi bhi trial ise nahi karwa sakta.
Sure event. 'Pointer 9 se chhote number par rukta hai' mein har outcome hai, isliye yeh khud $S$ hai, sure event: har trial ise karwata hai.
Simple event. 'Pointer 5 par rukta hai' hai ${5}$ . Jis event mein theek ek sample point ho, woh simple (ya elementary) event hai. $n$ outcomes wale sample space mein theek $n$ simple events hote hain; spinner ke aath hain, ${1}$ se ${8}$ tak.
Compound event. 'Pointer perfect square par rukta hai' hai ${1, 4}$ . Jis event mein ek se zyada sample points hon, woh compound event hai.

4/4'Perfect square' mein do outcomes: ${1, 4}$ .

Fig. 14.2Sample points ki ginti event ka type decide karti hai.NCERT p. 290

Ek spinner disc jo 1 se 8 tak numbered aath barabar sectors mein bata hai, aur right side ek text panel hai. Pehle step mein koi sector light nahi hai aur panel par likha hai 8 se zyada, empty set, impossible event. Doosre step mein aathon sectors light hain aur panel par likha hai 9 se kam, poora sample space, sure event. Teesre step mein sirf sector 5 light hai aur panel par likha hai 5 par rukta hai, 5 wala set, simple event. Chauthe step mein sectors 1 aur 4 light hain aur panel par likha hai perfect square, 1 aur 4 wala set, compound event.

Definition

Impossible aur sure events

Empty set $ϕ$ impossible event hai aur poora sample space $S$ sure event hai.

Definition

Simple aur compound events

Theek ek sample point wala event simple (elementary) hai; ek se zyada sample points wala event compound hai.

ExampleTeen coins toss karne par, 'teeno tails' $= {T T T}$ simple hai, jabki 'teeno faces ek jaise' $= {H H H, T T T}$ compound hai.

Quick check

3 mein se 1

Teen coins toss kiye jaate hain. Inmein se kaun sa event simple hai?

theek do headsteeno faces sameteeno tailskam se kam ek tail

Answer dekho

Answer (C) teeno tails

'Teeno tails' hai

{T T T},

ek akela sample point. 'Teeno faces same' hai

{H H H, T T T},

jismein do points hain, isliye woh compound hai.

14.1.3 Events ka algebra

Chapter 1 mein humne sets ko complement, union, intersection aur difference se joda tha. Events sets hi hain, isliye yahi operations purane events se naye events banate hain, aur har operation ka seedha-saadha matlab hai. Maan lo $A$ aur $B$ sample space $S$ wale ek experiment ke events hain.

Ek chalte example ke liye, saal ka ek mahina randomly chuno, to $S$ 12 mahinon ka set hai. Maan lo $A$ hai 'mahine mein 31 din hain' aur $B$ hai 'mahine ka naam J se shuru hota hai'. Tab $A$ = {Jan, Mar, May, Jul, Aug, Oct, Dec} aur $B$ = {Jan, Jun, Jul}.

'Not $A$ '. $S$ ke woh outcomes ikatthe karo jo $A$ ke bahar hain. Yeh complementary event $A^{'} = S - A$ banate hain, jo theek un trials mein hota hai jahan $A$ nahi hota. Yahan $A^{'}$ = {Feb, Apr, Jun, Sep, Nov}, yaani 31 din se chhote mahine.
' $A$ or $B$ '. Union $A \cup B$ mein woh outcomes hain jo $A$ mein, $B$ mein ya dono mein hain, isliye 'A or B' tab hota hai jab unmein se kam se kam ek hota hai. Yahan $A \cup B$ = {Jan, Mar, May, Jun, Jul, Aug, Oct, Dec}.
' $A$ and $B$ '. Intersection $A \cap B$ , yaani dono events ke common outcomes, tab hota hai jab dono hote hain. Yahan yeh {Jan, Jul} hai.
' $A$ but not $B$ '. Difference $A - B = A \cap B^{'}$ tab hota hai jab $A$ hota hai aur $B$ nahi. Yahan yeh {Mar, May, Aug, Oct, Dec} hai.

3/5 $A \cup B$ : kam se kam ek tag wala har mahina.

Fig. 14.3Ek sample space, A aur B se bane chaar naye events.NCERT p. 291

Baarah tiles, Jan se Dec, chaar-chaar ki teen rows mein. 31 din wale mahinon ki tiles par top left mein chhota tag A hai, aur jin mahinon ka naam J se shuru hota hai unpar top right mein tag B. Pehle step mein sirf tags dikhte hain. Doosre step mein not A ke liye bina tag A wali paanch tiles bhari jaati hain. Teesre step mein A or B ke liye tag A ya tag B wali har tile bhari jaati hai, aath mahine. Chauthe step mein A and B ke liye sirf Jan aur Jul, jinpar dono tags hain, bhare jaate hain. Paanchve step mein A but not B ke liye tag A wali lekin bina tag B wali paanch tiles bhari jaati hain.

Table 14.2Event ki bhasha aur set ki bhasha

Shabdon mein	Set ke roop mein
not $A$	$A^{'}$ , ya $S - A$
$A$ or $B$ (kam se kam ek)	$A \cup B$
$A$ and $B$ (dono)	$A \cap B$
$A$ but not $B$ (sirf $A$ )	$A - B = A \cap B^{'}$
na $A$ na $B$	$A^{'} \cap B^{'} = (A \cup B)^{'}$
$A$ , $B$ mein se theek ek	$(A - B) \cup (B - A)$

'Neither' wali row bhes badla hua De Morgan's law hai: 'not A and not B' wahi hai jo 'not (A or B)'. Yeh table paas rakho; zyaadatar worded problems har phrase ko iski kisi ek row mein translate karke solve hote hain.

Example 14.1NCERT Example 1 · p. 292

Divisors aur primes

1 se 12 tak ek number randomly chuna jaata hai. Maan lo $A$ hai 'number 12 ko divide karta hai' aur $B$ hai 'number prime hai'. In events ko sets ki tarah likho: (i) $A$ and $B$ , (ii) $A$ or $B$ , (iii) not $A$ , (iv) $A$ but not $B$ , (v) neither $A$ nor $B$ .

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

Quick check

3 mein se 1

Event '

A

aur

B

mein se theek ek hota hai' kaun sa set hai?

A \cup B

A^{'} \cap B^{'}

(A \cap B)^{'}

(A - B) \cup (B - A)

Answer dekho

Answer (D) $(A - B) \cup (B - A)$

Theek ek ka matlab hai bina

B

A

, ya bina

A

B

A \cup B

mein woh outcomes bhi hain jahan dono hote hain, aur

(A \cap B)^{'}

mein woh bhi hain jahan koi nahi hota.

14.1.4 Mutually exclusive events

52 cards ke achhi tarah shuffle kiye pack se ek card nikalo. Maan lo $A$ hai 'heart' aur $B$ hai 'black card'. Hearts laal hote hain, isliye koi card dono events mein nahi hai: inmein se jo bhi ho, doosra nahi ho sakta. Ab maan lo $C$ hai 'king'. King of hearts $A$ mein bhi hai aur $C$ mein bhi, isliye yeh do events saath mein ho sakte hain (Fig. 14.4).

3/3Hearts aur kings mein king of hearts common hai: $A \cap C \neq ϕ .$

Fig. 14.4Disjoint events saath mein nahi ho sakte; overlapping events ho sakte hain.NCERT p. 292

52 cards ek grid ki tarah bane hain, har suit ki ek row, spades, hearts, diamonds aur clubs, aur ace se king tak har rank ka ek column. Pehle step mein kuch mark nahi hai. Doosre step mein hearts wali row bhari hai aur do black rows, spades aur clubs, hatched hain; bhare aur hatched cells kabhi ek hi cell share nahi karte, aur label kehta hai koi common card nahi. Teesre step mein hearts wali row bhari hai aur king wala column outlined hai; king of hearts woh ek cell hai jo dono mein hai, star se marked, aur label kehta hai ki dono ek card share karte hain.

Definition

Mutually exclusive events

Events $A$ aur $B$ mutually exclusive (परस्पर अपवर्जी) hain jab unmein koi outcome common nahi hota, $A \cap B = ϕ,$ taaki koi bhi ek trial dono ko ek saath na karwa sake.

ExampleEk draw mein, 'heart' aur 'black card' mutually exclusive hain; 'heart' aur 'king' nahi hain.

Example 14.2NCERT Example 2 · p. 293

Kaun se pairs saath mein nahi ho sakte?

Do dice roll kiye jaate hain. Events lo $A$ : 'numbers ka product odd hai', $B$ : 'sum odd hai', $C$ : 'dono dice par same number hai' aur $D$ : 'product 30 se bada hai'. Events ke chhe pairs mein se kaun se kabhi saath mein nahi ho sakte?

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

36 outcomes par events ko haath se check karna dheema kaam hai. Neeche wala lab do dice ke sample space ko grid ki tarah dikhata hai, taaki tum ek nazar mein dekh sako ki do events overlap karte hain ya nahi, aur unka union, intersection aur complements kaise dikhte hain.

InteractiveDo dice ka event lab

Do dice roll karne ke 36 outcomes par ek event lab. Do drop-down menus baarah presets mein se events A aur B chunte hain, jaise sum even hai, double, ya pehle die par 6. Saat buttons dikhane wala event chunte hain: A, B, not A, A or B, A and B, A but not B, aur neither. Ek six by six grid, side mein pehla die aur upar doosra die, chune hue event ke cells bharta hai, aur har cell par chhote A aur B tags hote hain jab woh un events mein ho. Ek panel outcomes ki sankhya aur 36 ke fraction ke roop mein probability deta hai, addition rule check karta hai, aur batata hai ki A aur B mutually exclusive hain ya nahi aur milkar exhaustive hain ya nahi. Jab tak learner kuch nahi karta, kuch nahi hilta.

Fig. 14.5Events

A

aur

B

aur ek operation chuno. Grid us event ke outcomes ko bharta hai; panel unhe ginta hai aur batata hai ki

A

aur

B

mutually exclusive hain ya exhaustive.

14.1.5 Exhaustive events

Wapas aath sectors wale spinner par chalo. Maan lo $A$ hai 'zyada se zyada 3 wala number' $= {1, 2, 3}$ , $B$ hai 'even number' $= {2, 4, 6, 8}$ aur $C$ hai 'kam se kam 5 wala number' $= {5, 6, 7, 8} .$ 1 se 8 tak har number inmein se kam se kam ek mein hai, isliye $A \cup B \cup C = S$ : spinner kuch bhi dikhaye, teeno events mein se kam se kam ek hota hai. Is property wale events ko exhaustive kehte hain.

Definition

Exhaustive events

Events $E_{1}, E_{2}, \dots, E_{n}$ ki family exhaustive (निःशेष) hai jab woh milkar poora sample space cover kar lete hain: $E_{1} \cup E_{2} \cup \dots \cup E_{n} = S .$ Tab koi bhi trial aisa khatam nahi ho sakta ki inmein se koi bhi na ho.

Upar ke events $A$ , $B$ , $C$ overlap karte hain, kyunki $2 \in A \cap B$ aur $6 \in B \cap C,$ isliye yeh exhaustive hain lekin mutually exclusive nahi. Ab spinner ko alag tarah se baanto: $P$ = 'zyada se zyada 3', $Q$ = '4 ya 5' aur $R$ = 'kam se kam 6'. Yeh $S$ ko cover karte hain aur inmein se koi do ek outcome share nahi karte, isliye har spin par inmein se theek ek hota hai (Fig. 14.6). Aisi family ko mutually exclusive and exhaustive kehte hain; yeh sample space ko partition karti hai.

Mutually exclusive and exhaustive

\begin{aligned} E_{1} \cup \dots \cup E_{n} = S, \\ E_{i} \cap E_{j} = ϕ (i \neq j) \end{aligned}

(14.1)

jahan

$E_{1}, \dots, E_{n}$: family ke events
$i \neq j$: family ke koi bhi do alag events

2/3Har number cover hai, lekin 2, 6 aur 8 do baar cover hain.

Fig. 14.6Exhaustive: har column mein dot hai. Partition: har column mein theek ek.NCERT p. 293

Upar 1 se 8 tak aath numbered cells hain. Neeche teen rows hain, har event ki ek, aur event mein aane wale har number ke neeche ek dot. Pehle step mein rows hain A zyada se zyada 3, B even aur C kam se kam 5. Doosre step mein dots ke neeche ek count row aati hai: har number par kam se kam ek dot hai, isliye events exhaustive hain, lekin 2, 6 aur 8 par do-do dots hain, jo overlaps ki tarah marked hain. Teesre step mein rows badal kar P zyada se zyada 3, Q 4 ya 5, aur R kam se kam 6 ho jaati hain, aur ab har number par theek ek dot hai, jisse partition banta hai.

Example 14.3NCERT Example 3 · p. 294

Quiz mein guess karna

Ek student teen true-or-false questions ke answers guess karta hai, isliye har answer right (R) ya wrong (W) hai. Maan lo $A$ hai 'teeno right', $B$ hai 'theek do right' aur $C$ hai 'zyada se zyada ek right'. Kya $A$ , $B$ aur $C$ ek mutually exclusive and exhaustive family banate hain?

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

Outcomes ko kisi count ki value ke hisaab se baantna, jaisa yahan kiya, hamesha partition deta hai. Mahinon ko 31 din wale, 30 din wale aur February mein baantna ek aur example hai.

Quick check

3 mein se 1

Ek die roll kiya jaata hai. Kaun si family mutually exclusive aur exhaustive hai?

{1, 2}, {2, 3, 4}, {5, 6}

{1}, {2, 3}, {4, 5, 6}

{1, 3}, {2, 4}, {6}

{1, 2, 3}, {3, 4, 5}, {6}

Answer dekho

Answer (B) ${1}, {2, 3}, {4, 5, 6}$

{1}, {2, 3}, {4, 5, 6}

koi outcome share nahi karte aur milkar

S

dete hain. Pehli aur aakhri families overlap karti hain (2 par aur 3 par); teesri mein 5 chhoot gaya hai, isliye woh exhaustive nahi hai.

14.2 Probability ka axiomatic approach

Probably, likely aur hardly ever jaise shabd chances ko rank to karte hain, lekin naapte nahi. Probability theory har event ko ek number deti hai. Yeh number chunne ke do tareeke tum pehle se jaante ho. Classical tareeka favourable outcomes ki ginti ko equally likely outcomes ki ginti se divide karta hai. Experimental tareeka un trials ka fraction use karta hai jinmein event hua.

Axiomatic approach, jo Russian mathematician A. N. Kolmogorov ne 1933 mein diya, ek level upar kaam karta hai. Yeh nahi batata ki kaun se numbers use karne hain. Yeh teen rules list karta hai jinhe har samajhdaar assignment ko follow karna padta hai, aur baaki sab unhi se prove hota hai.

AxiomProbability ke axioms

Maan lo ek experiment ka sample space $S$ hai. $S$ par ek probability (प्रायिकता) ek rule $P$ hai jo har event $E \subseteq S$ ko ek real number $P (E)$ deta hai, teen properties ke saath. (i) Non-negativity: har event $E$ ke liye $P (E) \geq 0$ . (ii) Normalisation: sure event ko $P (S) = 1$ milta hai. (iii) Additivity: jab bhi $E \cap F = ϕ,$ tab $P (E \cup F) = P (E) + P (F) .$

Yaani $P$ ke inputs events hain, $S$ ke power set ke members, aur iske outputs real numbers hain. Axioms yeh nahi kehte ki yeh numbers 1 se neeche rehte hain; yeh ek consequence hai, jo neeche prove kiya gaya hai, ek aur aise consequence ke saath jo baar-baar use hota hai.

DerivationAxioms ke do jaldi wale consequences

Impossible event ki probability 0 hai. Kisi bhi event $E$ ke liye, sets $E$ aur $ϕ$ disjoint hain aur $E \cup ϕ = E .$ Axiom (iii) se $P (E) = P (E) + P (ϕ),$ milta hai, isliye $P (ϕ) = 0$ .

Koi probability 1 se zyada nahi hoti. $E$ aur $E^{'}$ disjoint hain aur $E \cup E^{'} = S$ , isliye axioms (iii) aur (ii) se $P (E) + P (E^{'}) = 1.$ milta hai. Axiom (i) se $P (E^{'}) \geq 0$ hai, isliye $P (E) \leq 1$ .

Jab sample space finite ho, $S = {ω_{1}, ω_{2}, \dots, ω_{n}},$ to axioms ek simple recipe ban jaate hain. $P ({ω_{i}}),$ yaani elementary event ${ω_{i}}$ ki probability, ke liye $P (ω_{i})$ likho. Tab har $i$ ke liye $0 \leq P (ω_{i}) \leq 1$ , aur

Outcome probabilities ka total 1 hota hai

\sum_{i = 1}^{n} P (ω_{i}) = 1

(14.2)

Event ki probability

P (A) = \sum_{ω_{i} \in A} P (ω_{i})

(14.3)

jahan

$A$: koi bhi event
$ω_{i} \in A$: sum sirf $A$ ke outcomes par chalta hai

Dono axiom (iii) se aate hain: simple events pairwise disjoint hain, isliye (iii) ko baar-baar lagane se unki probabilities jud jaati hain. Poore $S$ par total $P (S) = 1$ hai, jo (14.2) hai; $A$ ke outcomes par yeh $P (A)$ hai, jo (14.3) hai. Ulta bhi sach hai: $[0, 1]$ mein ke koi bhi numbers jinka total 1 ho, outcomes ko diye ja sakte hain, aur tab (14.3) ek aisa $P$ define karta hai jo teeno axioms follow karta hai.

Isliye axioms numbers fix nahi karte. Ek drawing pin zameen par giraao: woh ya to point upar karke ( $U$ ) girti hai ya side par tilt hokar ( $D$ ). Uske shape mein aisa kuch nahi jo dono ko equally likely banaye. $0 \leq p \leq 1$ wale har $p$ ke liye, choice $P (U) = p$ , $P (D) = 1 - p$ axioms follow karti hai, isliye infinitely many valid assignments hain. Asli pin ko kaun sa describe karta hai, yeh experiment ka sawaal hai. Coin ke liye bhi, $P (H) = 0.6$ , $P (T) = 0.4$ mathematically valid hai; yeh bas fair coin ki jagah ek muda hua coin describe karta hai.

SimulationUnequal sectors wala spinner

Teen unequal size ke sectors wala spinner lab: green disc ka aadha hissa cover karta hai, blue ek-tihaai aur red chhatha hissa. Buttons ise ek baar, 10, 100 ya 1000 baar ghumaate hain. Ek bar chart har colour ki relative frequency uski theoretical probability ke saath dikhata hai, aur ek readout green ki relative frequency track karta hai. Jab tak learner spin button nahi dabata, kuch nahi hilta.

Fig. 14.7Sectors disc ka

\frac{1}{2}

\frac{1}{3}

aur

\frac{1}{6}

hissa cover karte hain. Spin button dabao aur har bar ko uski theoretical value se compare karo.

Try karo

Spin karne se pehle: is disc ke liye assignment $P (Green), P (Blue), P (Red)$ likho aur use axioms se check karo.
1000 baar spin karo. Kya green ki relative frequency $\frac{1}{2}$ ke paas hai? Teeno relative frequencies jodo: tumhe hamesha kya total milta hai?
Sirf 10 spins ke baad bhi frequencies ka total 1 hota hai aur woh kabhi negative nahi hotin, phir bhi woh sector sizes se kaafi door ho sakti hain. Woh phir bhi kaun se axioms follow karti hain?

Example 14.4NCERT Example 4 · p. 297

Valid hai ya nahi?

Ek sample space hai $S = {ω_{1}, ω_{2}, ω_{3}, ω_{4}, ω_{5}} .$ Outcomes ko probabilities dene wale inmein se kaun se assignments valid hain?

Table 14.3Paanch proposed assignments

Row	$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$
(a)	0.1	0.3	0.2	0.25	0.15
(b)	0.5	0	0.5	0	0
(c)	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{16}$	$\frac{1}{16}$
(d)	0.3	0.3	0.3	0.3	−0.2
(e)	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$

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

14.2.1 Event ki probability

Jab har outcome ki probability tay ho jaaye, to (14.3) kisi bhi event ki probability deta hai: uske outcomes ki probabilities jod do. Outcomes ka equally likely hona zaroori nahi.

Ek aisa loaded die lo jismein 6 aane ki probability $\frac{1}{4}$ hai, aur baaki paanch faces bache hue $\frac{3}{4}$ ko barabar baantte hain, har ek $\frac{3}{20} = 0.15$ . Event 'even number' ke liye hum 2, 4 aur 6 ke hisse jodte hain: $0.15 + 0.15 + 0.25 = 0.55.$ 'Kam se kam 5' ke liye hum jodte hain $0.15 + 0.25 = 0.40.$ Fig. 14.8 total probability 1 ko in hisson mein kati ek strip ki tarah dikhata hai.

3/4P(even) $= 0.15 + 0.15 + 0.25 = 0.55.$

Fig. 14.8Total 1 ko outcomes mein baanto; event ko uske hisson ka sum milta hai.NCERT p. 298

Ek lambi horizontal strip total probability 1 ko dikhati hai, left end par 0 aur right end par 1. Pehle step mein yeh ek hi piece hai jis par label hai P of S barabar 1. Doosre step mein yeh 1 se 6 tak labelled chhe hisson mein kat jaati hai: 0.15 ke paanch hisse aur 0.25 ka ek lamba aakhri hissa. Teesre step mein 2, 4 aur 6 ke hisse strip ke upar uthte hain aur sum 0.15 plus 0.15 plus 0.25 barabar 0.55 dikhta hai, even number ki probability. Chauthe step mein iski jagah 5 aur 6 ke hisse uthte hain, aur kam se kam 5 ke liye 0.40 milta hai.

ExampleNCERT p. 298

Do matches, unequal chances

Ek analyst kisi team ke agle do matches ke results ko yeh probabilities deta hai, jahan W jeet hai aur L haar: $P (W W) = 0.36,$ $P (W L) = 0.24,$ $P (L W) = 0.24,$ $P (L L) = 0.16.$ Kya yeh assignment valid hai? Probability nikalo ki team (i) kam se kam ek baar jeetti hai, (ii) dono matches mein same result paati hai.

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

14.2.2 Equally likely outcomes ki probabilities

Aksar symmetry (fair coin, balanced die, achhi tarah shuffle kiya pack) saare outcomes ko equally likely bana deti hai. Maan lo $S = {ω_{1}, \dots, ω_{n}}$ aur har $i$ ke liye $P (ω_{i}) = p$ . Tab (14.2) kehta hai $n p = 1$ , isliye $p = \frac{1}{n} .$ Agar event $E$ mein $m$ outcomes hain, to (14.3) $\frac{1}{n}$ ki $m$ copies jodta hai:

Equally likely outcomes

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

(14.4)

jahan

$n (E) = m$: $E$ ke favourable outcomes ki sankhya
$n (S) = n$: outcomes ki total sankhya

Yeh Class 10 wala formula hai, jo ab axioms se derive hua hai, aur derivation iski limits bhi dikhata hai: yeh sirf tab sahi hai jab outcomes equally likely hon. Section 14.2.1 ke loaded die ke liye, $P (6) = \frac{1}{4},$ hai, $\frac{1}{6}$ nahi. Fair die ke saath toss kiye fair coin ke liye saare 12 outcomes equally likely hain, isliye $P (head and even) = \frac{3}{12} = \frac{1}{4} .$

Quick check

3 mein se 1

S = {a, b, c},

ke liye kaun sa assignment valid nahi hai?

0.7, 0.4, - 0.1

0.5, 0.5, 0

\frac{1}{3}, \frac{1}{3}, \frac{1}{3}

0.1, 0.2, 0.7

Answer dekho

Answer (A) $0.7, 0.4, - 0.1$

0.7 + 0.4 - 0.1 = 1,

lekin negative value axiom (i) todti hai. Baaki teen lists non-negative hain aur unka total 1 hai.

14.2.3 Event 'A or B' ki probability

1 se 20 tak numbered bees tokens mein se ek token randomly chuno. Maan lo $A$ hai 'even number' aur $B$ hai '3 ka multiple'. Tab $n (A) = 10$ aur $n (B) = 6$ , phir bhi $A \cup B$ mein 13 tokens hain, 16 nahi. Tokens 6, 12 aur 18 dono events mein hain, aur sum $n (A) + n (B)$ inmein se har ek ko do baar ginta hai (Fig. 14.9). Overlap ko ek baar ghataane se ginti theek ho jaati hai.

\begin{aligned} P (A \cup B) & = \frac{10 + 6 - 3}{20} \\ = \frac{13}{20} \end{aligned}

3/4Tiles 6, 12 aur 18 do baar gine gaye.

Fig. 14.9Dono counts jodne se overlap do baar gin jaata hai; use ek baar ghatao.NCERT Fig. 14.1 · p. 301

1 se 20 tak numbered bees tiles paanch-paanch ki chaar rows mein hain, neeche ek running count ke saath. Pehle step mein das even tiles bhari jaati hain aur unpar tag A lagta hai, aur count kehta hai n of A barabar 10. Doosre step mein 3 ke chhe multiples ko dashed border aur tag B milta hai, aur count kehta hai 10 plus 6 barabar 16. Teesre step mein tiles 6, 12 aur 18, jinpar dono tags hain, ko mota ring aur label do baar gina milta hai. Chauthe step mein A or B ke bahar wali tiles fade ho jaati hain, rings badal kar ek baar gina ho jaate hain, aur count kehta hai 16 minus 3 barabar 13, yaani A or B ka size.

Theorem 14.1Do events ka addition rule

Kisi bhi do events $A$ aur $B$ ke liye, unmein se kam se kam ek hone ki probability hai $P (A) + P (B)$ minus dono ke hone ki probability $P (A \cap B)$ . Symbols mein yeh neeche diya (14.5) hai.

Axioms se. $A \cup B$ aur $B$ ko disjoint pieces mein kaato: $A \cup B = A \cup (B - A)$ aur $B = (A \cap B) \cup (B - A) .$ Har union mein dono pieces koi outcome share nahi karte, isliye axiom (iii) dono par lagta hai.

$P (B - A)$ ke liye $x$ likho. Pehla union deta hai $P (A \cup B) = P (A) + x,$ aur doosra deta hai $P (B) = P (A \cap B) + x .$

Doosri equation ko pehli se ghatao. Unknown $x$ cancel ho jaata hai, aur $P (B)$ ko right-hand side le jaane par theek yahi rule milta hai.

Sample points se. Finite $S$ ke liye, (14.3) se $P (A) + P (B)$ ko $P (ω)$ ke sum ki tarah likho. $A - B$ aur $B - A$ ke points ek baar aate hain, lekin $A \cap B$ ke points do baar aate hain, har sum mein ek baar. $P (A \cap B)$ ki ek copy hata dene par $A \cup B$ ka har point ek baar gina jaata hai, jo $P (A \cup B)$ hai.

Addition rule

\begin{aligned} P (A \cup B) \\ = P (A) + P (B) \\ - P (A \cap B) \end{aligned}

(14.5)

Jab $A$ aur $B$ mutually exclusive hon, to $A \cap B = ϕ$ aur $P (ϕ) = 0$ , isliye rule chhota hokar $P (A \cup B) = P (A) + P (B),$ ban jaata hai, jo axiom (iii) hai. Shabdon mein: 'or' ki probabilities seedhe tabhi judti hain jab events overlap na kar sakein. Kyunki $P (A \cap B) \geq 0,$ isliye events ka har pair $P (A \cup B) \leq P (A) + P (B) .$ bhi satisfy karta hai.

Ise two-dice event lab mein try karo: do overlapping events aur operation 'A or B' chuno. Panel $n (A)$ aur $n (B)$ jodta hai, $n (A \cap B)$ ghataata hai aur theek $n (A \cup B)$ par pahunchta hai.

Quick check

3 mein se 1

P (A) = 0.45

P (B) = 0.3

aur

P (A \cap B) = 0.12.

P (A \cup B)

nikalo.

Answer dekho

Answer 0.63

0.45 + 0.3 - 0.12 = 0.63.

14.2.4 Event 'not A' ki probability

1 se 25 tak numbered pachchees tokens mein se ek token nikalo aur maan lo $A$ hai 'perfect square'. Tab $A = {1, 4, 9, 16, 25}$ aur $P (A) = \frac{5}{25} = \frac{1}{5} .$ Baaki 20 tokens $A^{'}$ banate hain, isliye $P (A^{'}) = \frac{20}{25} = \frac{4}{5},$ jo theek $1 - P (A)$ hai (Fig. 14.10). Yeh koi sanyog nahi hai.

3/3 $A$ aur $A^{'}$ milkar $S$ bharte hain: $0.2 + 0.8 = 1$ .

Fig. 14.10Ek event aur uska complement sample space ko baant dete hain, isliye unki probabilities ka total 1 hota hai.NCERT p. 301

Five by five grid mein pachchees numbered tiles, right side ek panel ke saath. Pehle step mein paanch perfect squares, 1, 4, 9, 16 aur 25, solid border ke saath bhare hain aur panel par likha hai A, 25 mein se 5, P of A barabar 0.2. Doosre step mein baaki bees tiles dashed border ke saath hatched hain aur panel par likha hai not A, 25 mein se 20, 0.8. Teesre step mein dono tarah ki tiles saath dikhti hain, har tile inmein se theek ek mein hai, aur panel par likha hai 0.2 plus 0.8 barabar 1.

Theorem 14.2Complement rule

Har event $A$ ke liye, $P (not A) = P (A^{'}) = 1 - P (A) .$

Complement rule

P (A^{'}) = 1 - P (A)

(14.6)

Yeh rule tab bahut kaam aata hai jab 'not $A$ ' ko ginna $A$ se aasaan ho, khaas kar kam se kam ek wale questions mein. Die do baar roll karo: 'koi six nahi' mein 36 mein se $5 \times 5 = 25$ outcomes hain, isliye P(kam se kam ek six) $= 1 - \frac{25}{36} = \frac{11}{36} .$

SimulationPoore pack se draw karna

52 ke standard pack wala card-drawing lab. Buttons ek card, 10, 100 ya 1000 cards draw karte hain, har baar poore reshuffle kiye pack se. Ek bar chart dikhata hai ki har suit kitni baar aaya, uske theoretical one quarter hisse ke saath, aur ek readout face card ki relative frequency ko theoretical value 3 over 13 se compare karta hai. Jab tak learner draw button nahi dabata, kuch nahi hilta.

Fig. 14.11Har draw poore, abhi shuffle kiye pack se ek card hai. Readout 'face card' track karta hai; iska complement baaki har draw hai.

Try karo

Draw karne se pehle: P(face card) aur P(face card nahi) kya hain?
1000 cards draw karo. Face-card frequency ko 1 se ghatao aur use apni P(face card nahi) wali value se compare karo.
Kya chaaron suit bars $\frac{1}{4}$ ke paas tik jaati hain? Kaun si do bars milkar P(red card) ka estimate deti hain?

Example 14.5NCERT Example 5 · p. 302

Pack se ek card

52 cards ke achhi tarah shuffle kiye pack se ek card randomly liya jaata hai, saare cards equally likely hain. Probability nikalo ki woh (i) face card hai, (ii) king nahi hai, (iii) red face card hai, (iv) na heart hai na face card.

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

Example 14.6NCERT Example 6 · p. 303

Laddoos ka jar

Ek jar mein same size ke 15 laddoo hain: 6 besan, 5 motichoor aur 4 coconut. Bina dekhe ek nikala jaata hai. Probability nikalo ki woh (i) besan hai, (ii) coconut hai, (iii) motichoor nahi hai, (iv) besan ya coconut hai.

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

Example 14.7NCERT Example 7 · p. 304

Do applicants

Meera aur Kabir ek hi summer internship ke liye apply karte hain. Meera ke select hone ki probability 0.3 hai, Kabir ke select hone ki 0.25, aur dono ke select hone ki 0.1. Probability nikalo ki (a) dono mein se koi select nahi hota, (b) unmein se kam se kam ek select nahi hota, (c) unmein se theek ek select hota hai.

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

Example 14.8NCERT Example 8 · p. 305

Team chunna

4 girls aur 3 boys mein se 3 ki ek quiz team randomly chuni jaati hai. Probability nikalo ki team mein (a) koi boy nahi hai, (b) theek ek boy hai, (c) kam se kam do boys hain.

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

Quick check

3 mein se 1

Agar

P (A) = \frac{3}{8},

P (not A)

kya hai?

Answer dekho

Answer 0.625

1 - \frac{3}{8} = \frac{5}{8} = 0.625.

Miscellaneous Examples

Aakhri examples is chapter ko Chapter 6 ki counting ke saath milaate hain. Plan hamesha same hai: ek equally likely sample space describe karo, use permutations ya combinations se gino, event gino, aur divide karo. Ginne se pehle koi symmetry dhoondho jo seedha answer de sakti ho.

3/3Har pair ke ek order mein A, B se pehle hai: $\frac{3}{6} = \frac{1}{2} .$

Fig. 14.12A aur B ko swap karne se orders ke pairs ban jaate hain, isliye theek aadhe orders mein A pehle aata hai.NCERT p. 308

Chhe cards letters A, B aur C ke chhe orders dikhate hain, do-do ki teen rows mein. Pehle step mein chhe orders bas list kiye gaye hain: left mein ABC, ACB aur CAB, right mein BAC, BCA aur CBA. Doosre step mein swap A and B label wala double arrow har row ke dono cards ko jodta hai, yeh dikhate hue ki har right card uska left partner hi hai jismein A aur B badal diye gaye. Teesre step mein har pair ka left card, jismein A, B se pehle aata hai, outlined aur ticked hai, aur label kehta hai 6 mein se 3, isliye P barabar aadha.

Example 14.9NCERT Example 9 · p. 308

Photo ki line

Paanch dost A, B, C, D aur E photo ke liye random order mein ek row mein khade hote hain, saare orders equally likely hain. Probability nikalo ki (i) A kahin B ke left mein khada hai, (ii) A, B aur C left se right isi order mein khade hain, zaroori nahi ki saath-saath, (iii) A aur B dono end wali jagah lete hain, (iv) A aur B ek doosre ke bagal mein khade hain.

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

Example 14.10NCERT Example 10 · p. 309

Haath mein aces

52 ke achhi tarah shuffle kiye pack se 5 cards ka hand baanta jaata hai. Probability nikalo ki usmein (i) koi ace nahi hai, (ii) theek teen aces hain, (iii) kam se kam teen aces hain.

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

Example 14.11NCERT Example 11 · p. 309

Teen events ka addition rule

Teen events ka addition rule, neeche (14.7), prove karo: kisi bhi events $A$ , $B$ , $C$ ke liye, $A \cup B \cup C$ ki probability teeno single probabilities ka sum hai, minus teeno pairwise intersections ki probabilities, plus $P (A \cap B \cap C) .$

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

Teen events ka addition rule

\begin{aligned} P (A \cup B \cup C) \\ = P (A) + P (B) \\ + P (C) \\ - P (A \cap B) \\ - P (B \cap C) \\ - P (A \cap C) \\ + P (A \cap B \cap C) \end{aligned}

(14.7)

Example 14.12NCERT Example 12 · p. 310

Teen medals

Chhe runners P, Q, R, S, T aur U bina kisi tie ke 100 m race daudte hain, finish ke saare orders equally likely hain. Probability nikalo ki (a) P, Q aur R kramashah gold, silver aur bronze jeette hain, (b) P, Q aur R teeno medals kisi bhi order mein jeette hain.

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

Quick check

2 mein se 1

Chaar dost, jinmein Asha aur Ben bhi hain, chaar chairs ki row mein randomly baithte hain. Probability nikalo ki Asha aur Ben ek doosre ke bagal mein baithte hain.

Answer dekho

Answer 0.5

Unhe ek block mein chipka do:

4! = 24

orders mein se

3! \times 2 = 12

, isliye

\frac{1}{2}

Practice · 15 questions

Practice questions answers ke saath (NCERT “Chapter 14 par Miscellaneous Exercise”)

1
NumericalMediumNCERT Miscellaneous Exercise, Q1
Ek carton mein 6 mango, 9 orange aur 15 litchi juice packs hain. Chaar packs randomly nikale jaate hain. Probability nikalo ki
1. chaaron orange hain;
  Answer dekho
  Answer 0.004597701149425287
  $\frac{^{9} C_{4}}{^{30} C_{4}} = \frac{126}{27 405} = \frac{2}{435} \approx 0.0046.$
2. kam se kam ek litchi hai.
  Answer dekho
  Answer 0.9501915708812261
  'Koi litchi nahi' ka complement: $1 - \frac{^{15} C_{4}}{^{30} C_{4}} = 1 - \frac{1365}{27 405} = \frac{248}{261} \approx 0.950.$
NumericalMediumNCERT Miscellaneous Exercise, Q2
52 ke achhi tarah shuffle kiye pack se paanch cards nikale jaate hain. 2 hearts aur 3 clubs milne ki probability nikalo.
Answer dekho
Answer 0.008583433373349339
$\frac{^{13} C_{2} \times^{13} C_{3}}{^{52} C_{5}} = \frac{78 \times 286}{2 598 960} = \frac{143}{16 660} \approx 0.0086.$
3
NumericalEasyNCERT Miscellaneous Exercise, Q3
Ek die ke teen faces par number 4, do faces par 5 aur ek face par 6 hai. Ise ek baar roll kiya jaata hai. Nikalo
1. P(5);
  Answer dekho
  Answer 0.3333333333333333
  $\frac{2}{6} = \frac{1}{3} .$
2. P(4 ya 6);
  Answer dekho
  Answer 0.6666666666666666
  $\frac{3 + 1}{6} = \frac{2}{3} .$
3. P(4 nahi).
  Answer dekho
  Answer 0.5
  $1 - \frac{3}{6} = \frac{1}{2} .$
4
NumericalHardNCERT Miscellaneous Exercise, Q4
School fete ke raffle mein 5000 tickets bikte hain aur unmein se 20 ek jaise prizes jeette hain. Chaar decimal places tak, kuch na jeetne ki probability nikalo agar tum kharidte ho
1. ek ticket;
  Answer dekho
  Answer 0.996
  $\frac{4980}{5000} = 0.996.$
2. do tickets;
  Answer dekho
  Answer 0.992
  $\frac{^{4980} C_{2}}{^{5000} C_{2}} = \frac{4980 \times 4979}{5000 \times 4999} \approx 0.9920.$
3. paanch tickets.
  Answer dekho
  Answer 0.9802
  $\frac{^{4980} C_{5}}{^{5000} C_{5}} \approx 0.9802.$
5
NumericalHardNCERT Miscellaneous Exercise, Q5
120 students, jinmein tum aur tumhara ek dost bhi hain, randomly 50 ke ek section aur 70 ke ek section mein baante jaate hain. Probability nikalo ki
1. tum dono same section mein ho;
  Answer dekho
  Answer 0.5098039215686274
  $\frac{50}{120} \cdot \frac{49}{119} + \frac{70}{120} \cdot \frac{69}{119} = \frac{7280}{14 280} = \frac{26}{51} \approx 0.510.$
2. tum alag sections mein ho.
  Answer dekho
  Answer 0.49019607843137253
  (a) ka complement: $\frac{25}{51} \approx 0.490.$
NumericalHardNCERT Miscellaneous Exercise, Q6
Chaar alag logon ko chaar letters likhe jaate hain aur chaar envelopes par address likhe jaate hain. Letters ko randomly envelopes mein daala jaata hai, har envelope mein ek. Ek ya zyada letters ke sahi envelope mein pahunchne ka chance kya hai?
Solution dikhao
Letters ko 1 se 4 kaho. Jin arrangements mein koi letter apni jagah par nahi hai, woh hain 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321: kul nau. Isliye P(kam se kam ek sahi) $= 1 - \frac{9}{24} = \frac{15}{24} = \frac{5}{8} .$
7
NumericalMediumNCERT Miscellaneous Exercise, Q7
$A$ aur $B$ aise events hain ki $P (A) = 0.62$ , $P (B) = 0.48$ aur $P (A \cap B) = 0.27.$ Nikalo
1. $P (A \cup B)$ ;
  Answer dekho
  Answer 0.83
  $0.62 + 0.48 - 0.27 = 0.83.$
2. $P (A^{'} \cap B^{'})$ ;
  Answer dekho
  Answer 0.17
  $1 - P (A \cup B) = 0.17.$
3. $P (A \cap B^{'})$ ;
  Answer dekho
  Answer 0.35
  $0.62 - 0.27 = 0.35.$
4. $P (B \cap A^{'})$ .
  Answer dekho
  Answer 0.21
  $0.48 - 0.27 = 0.21.$
NumericalMediumNCERT Miscellaneous Exercise, Q8
Chhe volunteers school ka blood-donation camp chalate hain. Unmein se ek ko press se baat karne ke liye randomly chuna jaata hai. Probability nikalo ki speaker female hai ya 25 se kam umar ka hai.
Table 14.4Volunteers
Naam Gender Age (saal)
Tanvi F 22
Arjun M 31
Farhan M 24
Gurpreet F 35
Lalit M 28
Neha F 19
Answer dekho
Answer 0.6666666666666666
Female: 3; 25 se kam: 3 (Tanvi, Farhan, Neha); dono: Tanvi aur Neha. Isliye $\frac{3 + 3 - 2}{6} = \frac{2}{3} .$
9
NumericalHardNCERT Miscellaneous Exercise, Q9
Digits 0, 2, 5, 6, 9 se 6000 se bade four-digit numbers randomly banaye jaate hain. Probability nikalo ki number 5 se divisible hai jab
1. digits repeat ho sakte hain;
  Answer dekho
  Answer 0.39759036144578314
  Saare: $2 \times 5^{3} - 1 = 249$ (6000 hatakar). 0 ya 5 par khatam hone wale: $2 \times 5^{2} \times 2 - 1 = 99.$ P $= \frac{99}{249} = \frac{33}{83} \approx 0.398.$
2. digits repeat nahi ho sakte.
  Answer dekho
  Answer 0.5
  Saare: $2 \times 4 \times 3 \times 2 = 48.$ 0 ya 5 par khatam hone wale: $2 \times 2 \times (3 \times 2) = 24.$ P $= \frac{1}{2}$ .
NumericalEasyNCERT Miscellaneous Exercise, Q10
Ek bicycle lock mein teen wheels hain, har ek par 0 se 9 tak digits, aur yeh teen alag digits ke ek sequence par khulta hai. Teen alag digits ka ek random guess ise kholne ki probability kya hai?
Answer dekho
Answer 0.001388888888888889
$^{10} P_{3} = 10 \times 9 \times 8 = 720$ sequences, ek sahi: $\frac{1}{720} \approx 0.0014.$
Quick answerMedium
Assertion (A): Kisi bhi do events ke liye, $P (A \cup B) \leq P (A) + P (B) .$ Reason (R): Events ke har pair ke liye $P (A \cap B) \geq 0$ .
Dono true hain aur R, A ko explain karta haiDono true hain lekin R, A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (A) Dono true hain aur R, A ko explain karta hai
(14.5) se, $P (A \cup B)$ hai $P (A) + P (B)$ mein se non-negative number $P (A \cap B)$ ghataaya hua, aur non-negative number ghataane se sum badh nahi sakta.
Quick answerMedium
Events $A$ , $B$ , $C$ mutually exclusive aur exhaustive hain, aur $P (A) = 2 P (B) = 3 P (C) .$ Tab $P (A)$ hai
$\frac{1}{2}$ $\frac{3}{11}$ $\frac{2}{3}$ $\frac{6}{11}$
Answer dekho
Answer (D) $\frac{6}{11}$
$P (A) + \frac{P (A)}{2} + \frac{P (A)}{3} = 1,$ isliye $\frac{11}{6} P (A) = 1$ aur $P (A) = \frac{6}{11} .$
Quick answerMedium
Do fair dice roll kiye jaate hain. Probability nikalo ki sum 8 hai ya dice par double aata hai.
Answer dekho
Answer 0.2777777777777778
Sum 8: 5 outcomes; doubles: 6; dono: $(4, 4)$ . Isliye $\frac{5 + 6 - 1}{36} = \frac{10}{36} = \frac{5}{18} \approx 0.278.$
Quick answerHard
$P (A) = 0.4$ aur $P (B) = 0.5$ . Kaun sa interval $P (A \cap B)$ ki har possible value rakhta hai, aur usse zyada kuch nahi?
$[0.1, 0.4]$ $[0, 0.5]$ $[0.2, 0.4]$ $[0, 0.4]$
Answer dekho
Answer (D) $[0, 0.4]$
$P (A \cap B) \leq P (A) = 0.4$ (jab $A \subseteq B$ ho tab yeh value milti hai). $P (A \cup B) \leq 1$ se humein sirf $P (A \cap B) \geq - 0.1,$ milta hai, isliye lower limit 0 hai (jab $A$ , $B$ mutually exclusive hon tab milti hai).
Quick answerHard
52 ke achhi tarah shuffle kiye pack se chaar cards nikale jaate hain. Probability nikalo ki chaaron alag suits ke hain.
Answer dekho
Answer 0.10549819927971188
$\frac{13^{4}}{^{52} C_{4}} = \frac{28 561}{270 725} \approx 0.1055.$

Naam	Gender	Age (saal)
Tanvi	F	22
Arjun	M	31
Farhan	M	24
Gurpreet	F	35
Lalit	M	28
Neha	F	19

Recap

Summary aur key points

Event sample space ka koi bhi subset hai; yeh tab hota hai jab outcome usmein ho (Section 14.1).
$ϕ$ impossible event hai aur $S$ sure event; simple events mein ek sample point hota hai, compound events mein zyada (Section 14.1.2).
'Not A' hai $A^{'} = S - A$ , 'A or B' hai $A \cup B$ , 'A and B' hai $A \cap B$ , 'A but not B' hai $A - B = A \cap B^{'}$ (Section 14.1.3).
$A$ aur $B$ mutually exclusive hain jab $A \cap B = ϕ$ (Section 14.1.4).
Events exhaustive hain jab unka union $S$ ho; agar woh pairwise disjoint bhi hon to $S$ ko partition karte hain ((14.1)).
Axioms: $P (E) \geq 0$ , $P (S) = 1$ , aur mutually exclusive $E$ , $F$ ke liye $P (E \cup F) = P (E) + P (F)$ ; isliye $P (ϕ) = 0$ (Section 14.2).
Finite $S$ ke liye, outcome probabilities $[0, 1]$ mein hoti hain aur unka total 1 hota hai ((14.2)).
Event ki probability uske outcomes ki probabilities ka sum hai ((14.3)).
$n$ equally likely outcomes ke saath, $P (E) = \frac{n (E)}{n (S)}$ ((14.4)).
Kisi bhi do events ke liye, $P (A)$ aur $P (B)$ jodo, phir $P (A \cap B)$ ghatao taaki overlap ek hi baar gina jaaye; mutually exclusive events ke liye kuch nahi ghataana (Theorem 14.1).
Complement rule $P (A^{'}) = 1 - P (A)$ 'kam se kam ek' tak pahunchne ka shortcut hai (Theorem 14.2).
Teen events ke liye, singles jodo, pairs ghatao aur triple wapas jodo ((14.7)).

Timeline

Historical note: jua ki tables se axioms tak

c. 2500 BCE
Mohenjo-daro mein mitti ke dice
Harappan sites par paki mitti ke cube jaise dice khudaai mein mile hain, isliye India mein chance ke games bahut purane hain.
1560s
Chance ke games par Cardano
Italian physician Gerolamo Cardano (1501–1576) ne Liber de Ludo Aleae likhi, dice ke mathematics par pehli kitaab. Yeh sirf 1663 mein chhapi.
1654
Pascal aur Fermat
Ek juaari ke dice aur beech mein ruke game ke baare mein sawaal Blaise Pascal tak pahunche, jinke Pierre de Fermat ke saath letters ne unhe solve kiya aur is subject ki neev rakhi.
1657
Huygens ki textbook
Christiaan Huygens (1629–1695) ne probability par pehla printed treatise publish kiya, jo expected value ke idea par based tha.
1713
Bernoulli ka law
Jacob Bernoulli ki Ars Conjectandi, jo unki maut ke baad publish hui, ne prove kiya ki bahut saare trials mein relative frequencies probabilities ke paas tik jaati hain.
1718–1812
De Moivre aur Laplace
Abraham de Moivre ki Doctrine of Chances aur Pierre-Simon Laplace ke 1812 ke treatise ne probability ko ek badi, systematic theory bana diya.
1800s
Russian school
P. L. Chebyshev (1821–1894) aur unke student A. A. Markov (1856–1922) ne general limit theorems prove kiye.
1933
Kolmogorov ke axioms
A. N. Kolmogorov (1903–1987) ne probability ko sets par ek function maana jo kuch axioms follow karta hai, yahi is chapter ka approach hai.

Flash cards · 6

Yaad rakhne wale rules

1 / 6

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 14.1 · Q1
NCERT Exercise 14.1, Q2	Practice 14.1 · Q2
NCERT Exercise 14.1, Q3	Practice 14.1 · Q3
NCERT Exercise 14.1, Q4	Practice 14.1 · Q4
NCERT Exercise 14.1, Q5	Practice 14.1 · Q5
NCERT Exercise 14.1, Q6	Practice 14.1 · Q6
NCERT Exercise 14.1, Q7	Practice 14.1 · Q7
NCERT Exercise 14.2, Q1	Practice 14.2 · Q1
NCERT Exercise 14.2, Q2	Practice 14.2 · Q2
NCERT Exercise 14.2, Q3	Practice 14.2 · Q3
NCERT Exercise 14.2, Q4	Practice 14.2 · Q4
NCERT Exercise 14.2, Q5	Practice 14.2 · Q5
NCERT Exercise 14.2, Q6	Practice 14.2 · Q6
NCERT Exercise 14.2, Q7	Practice 14.2 · Q7
NCERT Exercise 14.2, Q8	Practice 14.2 · Q8
NCERT Exercise 14.2, Q9	Practice 14.2 · Q9
NCERT Exercise 14.2, Q10	Practice 14.2 · Q10
NCERT Exercise 14.2, Q11	Practice 14.2 · Q11
NCERT Exercise 14.2, Q12	Practice 14.2 · Q12
NCERT Exercise 14.2, Q13	Practice 14.2 · Q13
NCERT Exercise 14.2, Q14	Practice 14.2 · Q14
NCERT Exercise 14.2, Q15	Practice 14.2 · Q15
NCERT Exercise 14.2, Q16	Practice 14.2 · Q16
NCERT Exercise 14.2, Q17	Practice 14.2 · Q17
NCERT Exercise 14.2, Q18	Practice 14.2 · Q18
NCERT Exercise 14.2, Q19	Practice 14.2 · Q19
NCERT Exercise 14.2, Q20	Practice 14.2 · Q20
NCERT Exercise 14.2, Q21	Practice 14.2 · Q21
NCERT Miscellaneous Exercise, Q1	Chapter 14 par Miscellaneous Exercise · Q1
NCERT Miscellaneous Exercise, Q2	Chapter 14 par Miscellaneous Exercise · Q2
NCERT Miscellaneous Exercise, Q3	Chapter 14 par Miscellaneous Exercise · Q3
NCERT Miscellaneous Exercise, Q4	Chapter 14 par Miscellaneous Exercise · Q4
NCERT Miscellaneous Exercise, Q5	Chapter 14 par Miscellaneous Exercise · Q5
NCERT Miscellaneous Exercise, Q6	Chapter 14 par Miscellaneous Exercise · Q6
NCERT Miscellaneous Exercise, Q7	Chapter 14 par Miscellaneous Exercise · Q7
NCERT Miscellaneous Exercise, Q8	Chapter 14 par Miscellaneous Exercise · Q8
NCERT Miscellaneous Exercise, Q9	Chapter 14 par Miscellaneous Exercise · Q9
NCERT Miscellaneous Exercise, Q10	Chapter 14 par Miscellaneous Exercise · Q10

Quick check: MCQs

14.1.2 Events ke types3 questions
14.1.3 Events ka algebra3 questions
14.1.5 Exhaustive events3 questions
14.2.2 Equally likely outcomes ki probabilities3 questions
14.2.3 Event 'A or B' ki probability3 questions
14.2.4 Event 'not A' ki probability3 questions
Miscellaneous Examples2 questions

Glossary · 12

Key terms (English + Hindi)

complementary eventपूरक घटनाpoorak ghatna: Event $A$ ke liye, $A$ mein na hone wale saare outcomes ka event $A^{'} = S - A$ ; yeh theek tab hota hai jab $A$ nahi hota.Chapter mein pehli baar yahan aaya →
compound eventमिश्र घटनाmishr ghatna: Ek se zyada sample points wala event.Chapter mein pehli baar yahan aaya →

equally likely outcomesसम-संभाव्य परिणामsam-sambhavya parinaam: Aise outcomes jinki probability same ho, $n$ outcomes hon to har ek ki $\frac{1}{n}$ .Chapter mein pehli baar yahan aaya →
eventघटनाghatna: Sample space ka koi bhi subset. Yeh tab hota hai jab experiment ka outcome usmein ho.
exhaustive eventsनिःशेष घटनाएँnihshesh ghatnaayen: Aise events jinka union poora sample space hai, isliye har trial mein unmein se kam se kam ek hota hai.Chapter mein pehli baar yahan aaya →

impossible eventअसंभव घटनाasambhav ghatna: Empty set $ϕ$ ko event ki tarah dekhna: koi outcome ise nahi karwata.Chapter mein pehli baar yahan aaya →

mutually exclusive eventsपरस्पर अपवर्जी घटनाएँparaspar apavarjee ghatnaayen: Aise events jo saath mein nahi ho sakte: $A \cap B = ϕ$ .

probabilityप्रायिकताpraayikta: Ek function $P$ jo har event ko ek number $P (E)$ deta hai, jahan $P (E) \geq 0$ , $P (S) = 1$ aur mutually exclusive $E$ , $F$ ke liye $P (E \cup F) = P (E) + P (F)$ .Chapter mein pehli baar yahan aaya →

random experimentयादृच्छिक परीक्षणyadrichchhik pareekshan: Aisa experiment jiske ek se zyada possible outcomes hon, sab pehle se pata hon, lekin asli outcome predict na kiya ja sake.Chapter mein pehli baar yahan aaya →

sample spaceप्रतिदर्श समष्टिpratidarsh samashti: Random experiment ke saare possible outcomes ka set $S$ ; har element ek sample point hai.Chapter mein pehli baar yahan aaya →
simple eventसरल घटनाsaral ghatna: Theek ek sample point wala event, jaise ${5}$ ; ise elementary event bhi kehte hain.Chapter mein pehli baar yahan aaya →
sure eventनिश्चित घटनाnishchit ghatna: Poore sample space $S$ ko event ki tarah dekhna: yeh har trial mein hota hai.Chapter mein pehli baar yahan aaya →

Probability: Class 11 Maths Chapter 14 Hinglish mein

Is chapter mein aap kya seekhenge

14.1 Event

14.1.1 Event ka hona (occurrence)

14.1.2 Events ke types

14.1.3 Events ka algebra

14.1.4 Mutually exclusive events

14.1.5 Exhaustive events

Practice 14.1: answers ke saath

14.2 Probability ka axiomatic approach

14.2.1 Event ki probability

14.2.2 Equally likely outcomes ki probabilities

14.2.3 Event 'A or B' ki probability

14.2.4 Event 'not A' ki probability

Practice 14.2: answers ke saath

Miscellaneous Examples

Practice questions answers ke saath (NCERT “Chapter 14 par Miscellaneous Exercise”)

Summary aur key points

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye