Probability: Class 12 Maths Chapter 13 Hinglish mein

Kuch naya pata chalne par chances ko update karna

Mathematics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

100 min
1 simulation
66 practice questions
21 MCQs

Is chapter mein aap kya seekhenge

Sample space ko F tak chhota karke, ya P(E ∩ F)/P(F) se, conditional probability P(E|F) nikaalo aur uski properties use karo
Multiplication rule se nikaalo ki do ya teen events sab ke sab hone ki probability kitni hai, without replacement draws ke saath bhi
Tay karo ki do ya teen events independent hain ya nahi, aur independent aur mutually exclusive events ko alag-alag samjho
Sample space ka partition pehchaano aur theorem of total probability use karo
Bayes' theorem se prior probability ko posterior probability mein update karo, trees aur natural frequencies se check karte hue
Random variable ko sample space par defined real-valued function ki tarah samjhao aur uski values list karo

NCERT Mathematics Class 12, Chapter 13, pp. 406–438 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

13.1 Introduction

Kya inter-house sports meet wale din baarish hogi? Kya 8:15 ki local time par Dadar pahunchegi? Class 9 se tum aise sawaalon par 0 aur 1 ke beech ek number lagaate aa rahe ho. Class 11 mein tumne ise do tarah se calculate kiya: equally likely outcomes gin kar, aur Kolmogorov ke axioms se, jo probability ko events par defined ek function maante hain. Tumne addition rule bhi dekha, jo $P (A \cup B)$ ko $P (A) + P (B)$ minus overlap $P (A \cap B)$ ke roop mein deta hai.

Asli faisle shayad hi kabhi zero information se shuru hote hain. Baadal ghirne par forecaster baarish ka chance badal deta hai; doctor bimaari ka naam lene se pehle test report ko tolta hai. Yeh chapter probability ke baare mein hai jab hum kuch naya jaante hain: conditional probability, multiplication rule, independence, aur Bayes' theorem, jo effect se uske likely cause tak ulta reasoning karta hai. Aakhir mein random variables par ek pehli nazar hai. Jab tak problem kuch aur na kahe, har experiment ke outcomes equally likely maane jaate hain.

Do dice se warm up karo. $(1, 1)$ se $(6, 6)$ tak ke ordered pairs 36 equally likely outcomes dete hain, aur unmein se chhah ka total 9 se zyada hai. Isliye total 9 se zyada hone ki classical probability $\frac{6}{36} = \frac{1}{6}$ hai. Neeche wala lab tumhare liye dice roll karta hai, taaki tum dekh sako ki relative frequency us value ke paas kaise aati hai.

SimulationDo dice: 9 se zyada totals

Ek two-dice probability lab. Buttons pair ko ek baar, 10, 100 ya 1000 baar roll karte hain. Ek bar chart dikhata hai ki 2 se 12 tak har total kitni baar aaya, uski theoretical probability ke saath, aur ek readout 9 se zyada total ki relative frequency ko theoretical value one sixth se compare karta hai. Jab tak learner roll button nahi dabata, kuch nahi hilta.

Fig. 13.1Har bar 2 se 12 tak ka ek total ginta hai. Track kiya gaya event 10, 11 ya 12 ka total hai, jiski theoretical probability

\frac{1}{6}

hai.

Try karo

Ek baar mein 100 roll karo jab tak lagbhag 1000 rolls na ho jaayein. Kya 9 se zyada totals ka share $\frac{1}{6} \approx 0.17$ ke kareeb hai?
Kaunsa total sabse zyada baar aata hai? Gino ki 36 ordered pairs mein se kitne use dete hain.
Section 13.2 se pehle: jin rolls ka total 9 se zyada hai, unmein se kitne fraction mein 6 hone ki ummeed hai?

Quick check

2 mein se 1

Do fair dice roll kiye jaate hain. Total theek 9 aane ki probability kya hai? (1/5 jaisa fraction chalega.)

Answer dekho

Answer 0.1111

Chaar ordered pairs 9 dete hain:

(3, 6), (4, 5), (5, 4), (6, 3)

. Isliye probability

\frac{4}{36} = \frac{1}{9}

hai.

(4, 5)

aur

(5, 4)

ko ek hi outcome ginne se denominator galat ho jaata.

13.2 Conditional Probability

Ek board game mein tumhe do dice mein se kisi par bhi 6 chahiye. Dice abhi cup ke neeche hain, tabhi ek dost jhaank kar kehta hai, total kam se kam 10 hai. Kya yeh achhi khabar hai? Hint se pehle kam se kam ek 6 ka chance $\frac{11}{36}$ hai, ek-tihaai se thoda kam. Hint ke baad sirf chhah outcomes bachte hain: $(4, 6)$ , $(5, 5)$ , $(5, 6)$ , $(6, 4)$ , $(6, 5)$ aur $(6, 6)$ . Inmein se paanch mein 6 hai, isliye chance uchhal kar $\frac{5}{6}$ ho gaya (Fig. 13.2).

2/3Event $F$ , 10 ya usse zyada ka total, sirf 6 possible outcomes chhodta hai.

Fig. 13.2Yeh jaanne par ki total kam se kam 10 hai, sample space 36 outcomes se ghat kar 6 reh jaata hai.

36 cells ka ek grid, throws ke har ordered pair ke liye ek cell, pehla die side mein neeche ki taraf aur doosra die upar ki taraf; har cell total dikhata hai. Pehle step mein 6 wale 11 cells, yaani aakhri row aur aakhri column, gol ghere gaye hain, aur panel par likha hai E, a 6 shows, 11 of 36 cells. Doosre step mein 10, 11 ya 12 total wale chhah cells, neeche right corner mein, outline kiye gaye hain aur baaki cells fade ho jaate hain; panel kehta hai ki ab sirf 6 cells possible hain. Teesre step mein baaki cells lagbhag gayab ho jaate hain, 6 cells ka naya sample space bachta hai, aur unmein se 5 ghere gaye hain, isliye conditional probability five sixths hai.

Chalo events ko naam dein. $E$ hai kam se kam ek die par 6 aur $F$ hai total 10 ya usse zyada. Jab hum jaante hain ki $F$ ho chuka hai, to $F$ ke bahar ka har outcome hat jaata hai, aur $F$ khud sample space ban jaata hai. $E$ ke jo outcomes bachte hain, woh theek $E \cap F$ ke outcomes hain. Is nayi probability ko hum $P (E ∣ F)$ likhte hain, aur padhte hain $F$ diya hone par $E$ ki probability.

\begin{aligned} P (E ∣ F) \\ = \frac{n (E \cap F)}{n (F)} = \frac{5}{6} \end{aligned}

Upar aur neeche dono ko $n (S) = 36$ se divide karo, to counts probabilities ban jaate hain: $\frac{5 / 36}{6 / 36}$ . Is doosre form ko equally likely outcomes ki zaroorat nahi, isliye yahi definition ka kaam karta hai.

Definition

Conditional probability

Ek hi sample space ke do events $E$ aur $F$ lo, jahan $P (F) \neq 0$ ho. $F$ diya hone par $E$ ki conditional probability, $F$ ki probability ka woh hissa hai jo $E$ mein bhi aata hai: $P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$ .

ExampleDo dice: $P (a 6 ∣ total \geq 10) = \frac{5 / 36}{6 / 36} = \frac{5}{6}$ .

Conditional probability

\begin{aligned} P (E ∣ F) \\ = \frac{P (E \cap F)}{P (F)} \\ agar P (F) \neq 0 \end{aligned}

(13.1)

jahan

$P (E \cap F)$: $E$ aur $F$ dono hone ki probability
$P (F)$: condition $F$ ki probability

$P (F) \neq 0$ par itna zor kyun? Impossible event par condition lagane se $\frac{0}{0}$ milta hai, jiski koi value nahi. Yeh bhi dhyaan do ki condition bar ke right mein baithti hai aur denominator badalti hai. Yahan $P (F ∣ E) = \frac{5 / 36}{11 / 36} = \frac{5}{11}$ hai, jo $P (E ∣ F) = \frac{5}{6}$ se alag number hai.

13.2.1 Conditional probability ki properties

Ek event $F$ fix karo jiske liye $P (F) \neq 0$ ho. Tab rule $E \mapsto P (E ∣ F)$ chhote sample space $F$ par ek normal probability ki tarah behave karta hai. Teen properties ise precise banati hain.

PropertyProperty 1: condition certain hai

$P (S ∣ F) = P (F ∣ F) = 1$ .

PropertyProperty 2: F diya hone par unions

Kisi bhi events $A$ aur $B$ ke liye, $P (A \cup B ∣ F)$ barabar hai $P (A ∣ F) + P (B ∣ F)$ minus $P (A \cap B ∣ F)$ ke. Jab $A$ aur $B$ disjoint hon, to aakhri term zero hota hai.

PropertyProperty 3: F diya hone par complements

$P (E^{'} ∣ F) = 1 - P (E ∣ F)$ .

Ek aur nateeja: $E \cap F$ , $F$ ka hissa hai, isliye $P (E \cap F) \leq P (F)$ , aur isliye $0 \leq P (E ∣ F) \leq 1$ , jaisa har probability ko hona chahiye.

Example 1NCERT Example 1 · p. 409

Formula ko dono taraf se padhna

Events $A$ aur $B$ ke liye $P (A) = 0.45$ , $P (B) = 0.6$ aur $P (A \cap B) = 0.24$ hai. $P (A ∣ B)$ aur $P (B ∣ A)$ nikaalo.

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

Example 2NCERT Example 2 · p. 409

Teen bachche, kam se kam do ladkiyan

Ek family mein teen bachche hain. Maan lo har bachcha ladki ya ladka hone ke equally likely hai, baaki bachchon se independently. Yeh diya hai ki kam se kam do bachche ladkiyan hain; teeno ke ladki hone ki probability nikaalo.

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

Example 3NCERT Example 3 · p. 410

Ek token jo perfect square hai

1 se 30 tak numbered tokens ek lucky draw ke liye bag mein mila diye gaye hain, aur ek token random se nikala jaata hai. Woh perfect square nikalta hai. Uske odd hone ki probability kya hai?

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

Example 4NCERT Example 4 · p. 410

Class 12 ke bus riders

Ek school mein 1200 students hain. Unmein se 520 school bus se aate hain, aur bus riders mein se 156 Class 12 mein hain. Random se chuna gaya ek student bus rider nikalta hai. Student ke Class 12 mein hone ki probability nikaalo.

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

Example 5NCERT Example 5 · p. 411

Teen tails ke baad

Ek fair coin chaar baar uchhala jaata hai. Pehle teen tosses mein sab tails aate hain. Chauthe toss mein head aane ki probability nikaalo.

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

Example 6NCERT Example 6 · p. 411

12 ka product

Ek die do baar phenka jaata hai aur dono numbers ka product 12 hai. Kisi ek throw mein 6 aane ki probability kya hai?

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

Ab tak har outcome equally likely tha. Definition $P (E \cap F) / P (F)$ tab bhi utni hi achhi chalti hai jab woh equally likely na hon; tab hum outcomes ginne ki jagah unki probabilities jodte hain. Jo experiments stages mein hote hain, unhe tree diagram se dikhana sabse achha hai: har stage ke har result ke liye ek branch, jis par usse pehle ki sab baaton ke diye hone par uski probability likhi hoti hai. Ek path ke saath multiply karne se us path ki probability milti hai (Fig. 13.3).

3/3No head ke paas probability ka $\frac{2}{3}$ bachta hai; do 6-paths uska $\frac{1}{6}$ rakhte hain.

Fig. 13.3Stages wala experiment ek tree ki tarah: path ke saath multiply karo, phir jo paths chahiye unhe jodo.NCERT Fig. 13.2 · p. 412

Ek tree jo left mein ek dot se shuru hota hai. Teen branches die ke results 6, 5 aur 1 to 4 tak jaati hain, probabilities one sixth, one sixth aur two thirds ke saath. 6 aur 5 se branches red (one third) aur green (two thirds) tak jaati hain; 1 to 4 se branches head aur tail tak jaati hain, dono one half. Doosre step mein har path ki probability uske leaf ke paas dikhti hai: one eighteenth, one ninth, one eighteenth, one ninth, one third aur one third. Teesre step mein head leaf fade hota hai, bina head wale paanch leaves outline hote hain, ek brace 6 ke baad wale do leaves ko E aur F dono ke roop mein mark karta hai, aur result likha hai P of E given F equals one quarter.

Example 7NCERT Example 7 · p. 412

Ek die, phir ball ya coin

Meera ek die phenkti hai. Agar 5 ya 6 aaye, to woh 1 red aur 2 green balls wale bag se ek ball nikaalti hai; warna woh coin uchhalti hai. Yeh diya hai ki koi head nahi aaya; die par 6 aane ki probability nikaalo.

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

Quick check

4 mein se 1

Do dice roll kiye jaate hain aur total 7 hai. Kisi ek die par 2 aane ki probability kya hai?

\frac{1}{6}

\frac{11}{36}

\frac{1}{3}

\frac{1}{18}

Answer dekho

Answer (C) $\frac{1}{3}$

Total 7 ke chhah outcomes hain, aur unmein se do,

(2, 5)

aur

(5, 2)

, mein 2 hai:

\frac{2}{6} = \frac{1}{3}

. Value

\frac{1}{18}

P (F)

se divide karne se pehle ka

P (E \cap F)

hai, aur

\frac{11}{36}

condition ko ignore karta hai.

Practice · 17 questions

Practice 13.1: answers ke saath

Short answerEasyNCERT Exercise 13.1, Q1
Events $E$ aur $F$ ke liye $P (E) = 0.5$ , $P (F) = 0.4$ aur $P (E \cap F) = 0.15$ hai. $P (E ∣ F)$ aur $P (F ∣ E)$ nikaalo.
Solution dikhao
Answer $\frac{3}{8}$ aur $0.3$
$P (E ∣ F) = \frac{0.15}{0.4} = 0.375 = \frac{3}{8}$ .
$P (F ∣ E) = \frac{0.15}{0.5} = 0.3$ .
Quick answerEasyNCERT Exercise 13.1, Q2
Agar $P (B) = 0.8$ aur $P (A \cap B) = 0.36$ ho, to $P (A ∣ B)$ nikaalo.
Answer dekho
Answer 0.45
$P (A ∣ B) = \frac{0.36}{0.8} = 0.45$ .
3
Short answerMediumNCERT Exercise 13.1, Q3
$P (A) = 0.6$ , $P (B) = 0.45$ aur $P (B ∣ A) = 0.5$ hai.
1. $P (A \cap B)$ nikaalo.
  Answer dekho
  Answer 0.3
  $P (A)$ ko $P (B ∣ A)$ se multiply karo: $0.6 \times 0.5 = 0.3$ .
2. $P (A ∣ B)$ nikaalo.
  Answer dekho
  Answer 0.6667
  $P (A ∣ B) = \frac{0.3}{0.45} = \frac{2}{3}$ .
3. $P (A \cup B)$ nikaalo.
  Answer dekho
  Answer 0.75
  $P (A \cup B) = 0.6 + 0.45 - 0.3 = 0.75$ .
Short answerMediumNCERT Exercise 13.1, Q4
Agar $3 P (A) = P (B) = \frac{3}{5}$ aur $P (A ∣ B) = \frac{1}{4}$ ho, to $P (A \cup B)$ nikaalo.
Solution dikhao
Answer $\frac{13}{20}$
$P (B) = \frac{3}{5}$ aur $P (A) = \frac{1}{5}$ .
$P (A \cap B) = P (B) P (A ∣ B) = \frac{3}{5} \cdot \frac{1}{4} = \frac{3}{20}$ .
Isliye $P (A \cup B)$ hai $\frac{4}{20} + \frac{12}{20} - \frac{3}{20} = \frac{13}{20}$ .
Short answerMediumNCERT Exercise 13.1, Q5
$P (A) = \frac{7}{15}$ , $P (B) = \frac{2}{5}$ aur $P (A \cup B) = \frac{3}{5}$ hai. Nikaalo (a) $P (A \cap B)$ , (b) $P (A ∣ B)$ , (c) $P (B ∣ A)$ .
Solution dikhao
Answer $\frac{4}{15}$ , $\frac{2}{3}$ , $\frac{4}{7}$
(a) Addition rule ko ulta padho: overlap $\frac{7}{15} + \frac{6}{15} - \frac{9}{15} = \frac{4}{15}$ hai.
(b) $P (A ∣ B) = \frac{4 / 15}{6 / 15} = \frac{2}{3}$ .
(c) $P (B ∣ A) = \frac{4 / 15}{7 / 15} = \frac{4}{7}$ .
6
Short answerMediumNCERT Exercise 13.1, Q6
Ek coin chaar baar uchhala jaata hai. Har case mein $P (E ∣ F)$ nikaalo.
1. $E$ : chauthe toss par tail; $F$ : pehle teeno tosses mein se har ek par tail.
  Answer dekho
  Answer 0.5
  $F = {T T T H, T T T T}$ aur $E \cap F = {T T T T}$ , isliye $P (E ∣ F) = \frac{1}{2}$ .
2. $E$ : theek do heads; $F$ : kam se kam ek head.
  Answer dekho
  Answer 0.4
  $F$ mein $16 - 1 = 15$ outcomes hain. Theek do heads unmein se $(\binom{4}{2}) = 6$ mein aate hain, isliye $P (E ∣ F) = \frac{6}{15} = \frac{2}{5}$ .
3. $E$ : kam se kam teen heads; $F$ : pehle toss par head.
  Answer dekho
  Answer 0.5
  Pehla head diya hai, to aakhri teen tosses mein kam se kam do heads chahiye: 8 mein se $3 + 1 = 4$ outcomes, isliye $P (E ∣ F) = \frac{1}{2}$ .
7
Short answerEasyNCERT Exercise 13.1, Q7
Ek coin uchhala jaata hai aur ek die phenka jaata hai. $P (E ∣ F)$ nikaalo.
1. $E$ : die par 6; $F$ : coin par head.
  Answer dekho
  Answer 0.16667
  $F$ mein chhah outcomes $(H, 1), \dots, (H, 6)$ hain aur unmein se ek mein 6 hai: $\frac{1}{6}$ .
2. $E$ : head aur 4 se bada number; $F$ : coin par tail.
  Answer dekho
  Answer 0
  $E$ ko head chahiye aur $F$ kehta hai tail, isliye $E \cap F = \emptyset$ aur $P (E ∣ F) = 0$ .
Quick answerEasyNCERT Exercise 13.1, Q8
Ek die chaar baar phenka jaata hai. $E$ : chauthe throw par 4 se zyada. $F$ : pehle teeno throws mein se har ek par 1. $P (E ∣ F)$ nikaalo.
Answer dekho
Answer 0.3333
$F$ pehle teen throws fix kar deta hai, chauthe ke liye 6 equally likely values bachti hain; unmein se do (5, 6) 4 se zyada hain. Isliye $P (E ∣ F) = \frac{2}{6} = \frac{1}{3}$ .
Short answerMediumNCERT Exercise 13.1, Q9
Chaar dost, Asha, Bina, Charu aur Dev, chaar seats ki ek row mein random se baithte hain. $E$ : Asha kisi ek end par baithti hai. $F$ : Bina doosri seat par baithti hai. $P (E ∣ F)$ nikaalo.
Solution dikhao
Answer $\frac{2}{3}$
$F$ diya hone par baaki teen dost seats 1, 3 aur 4 ko $3! = 6$ equally likely tareekon se bharte hain.
Unmein se $2 \times 2! = 4$ mein Asha end par (seat 1 ya seat 4) hai.
Isliye $P (E ∣ F) = \frac{4}{6} = \frac{2}{3}$ .
10
Short answerMediumNCERT Exercise 13.1, Q10
Ek white die aur ek blue die saath mein roll kiye jaate hain.
1. White die par 4 hai, yeh diya hone par total kam se kam 8 hone ki probability nikaalo.
  Answer dekho
  Answer 0.5
  Blue die par 4, 5 ya 6 aana chahiye: $\frac{3}{6} = \frac{1}{2}$ .
2. Blue die par odd number hai, yeh diya hone par total 5 hone ki probability nikaalo.
  Answer dekho
  Answer 0.1111
  $F$ mein $3 \times 6 = 18$ outcomes hain. Total 5 ke liye white 4 ke saath blue 1, ya white 2 ke saath blue 3 chahiye (blue 5 ke liye white 0 chahiye hota). Isliye $P = \frac{2}{18} = \frac{1}{9}$ .
Short answerMediumNCERT Exercise 13.1, Q11
Ek fair die phenka jaata hai. Maan lo $A = {2, 4, 6}$ , $B = {1, 2, 3, 4}$ aur $C = {3, 4, 5}$ . Nikaalo (a) $P (A ∣ B)$ aur $P (B ∣ A)$ , (b) $P (A ∣ C)$ aur $P (C ∣ A)$ , (c) $P (A \cup B ∣ C)$ aur $P (A \cap B ∣ C)$ .
Solution dikhao
Answer(a) $\frac{1}{2}$ aur $\frac{2}{3}$ (b) $\frac{1}{3}$ aur $\frac{1}{3}$ (c) $\frac{2}{3}$ aur $\frac{1}{3}$
(a) $A \cap B = {2, 4}$ : $P (A ∣ B) = \frac{2}{4} = \frac{1}{2}$ aur $P (B ∣ A) = \frac{2}{3}$ .
(b) $A \cap C = {4}$ : $P (A ∣ C) = \frac{1}{3}$ aur $P (C ∣ A) = \frac{1}{3}$ .
(c) $A \cup B = {1, 2, 3, 4, 6}$ , $C$ se ${3, 4}$ mein milta hai, isliye $P (A \cup B ∣ C) = \frac{2}{3}$ . $A \cap B = {2, 4}$ , $C$ se ${4}$ mein milta hai, isliye $P (A \cap B ∣ C) = \frac{1}{3}$ .
12
Short answerEasyNCERT Exercise 13.1, Q12
Ek cricket team apne agle do matches mein se har ek barabar chance se jeetti ya haarti hai, independently (koi tie nahi). Dono jeetne ki probability nikaalo, yeh diya hone par ki
1. woh pehla match jeetti hai;
  Answer dekho
  Answer 0.5
  ${W W, W L}$ diya hai, to sirf $W W$ chalta hai: $\frac{1}{2}$ .
2. woh kam se kam ek match jeetti hai.
  Answer dekho
  Answer 0.3333
  ${W W, W L, L W}$ diya hai, to sirf $W W$ chalta hai: $\frac{1}{3}$ .
Quick answerEasyNCERT Exercise 13.1, Q13
Ek school library mein 240 Hindi fiction books, 160 Hindi non-fiction books, 360 English fiction books aur 440 English non-fiction books hain. Random se ek book uthayi jaati hai aur woh English mein hai. Uske fiction hone ki probability kya hai?
Answer dekho
Answer 0.45
$360 + 440 = 800$ English books hain aur unmein se 360 fiction hain: $\frac{360}{800} = \frac{9}{20} = 0.45$ .
Quick answerMediumNCERT Exercise 13.1, Q14
Do dice phenke jaate hain aur dono numbers alag hain. Unka product 12 hone ki probability nikaalo.
Answer dekho
Answer 0.1333
Alag numbers: $36 - 6 = 30$ outcomes. Product 12: $(2, 6), (6, 2), (3, 4), (4, 3)$ , sab mein alag numbers. Isliye $P = \frac{4}{30} = \frac{2}{15}$ .
Short answerHardNCERT Exercise 13.1, Q15
Ek die phenka jaata hai. Agar even number aaye, to use phir se phenka jaata hai; agar odd number aaye, to coin uchhala jaata hai. Pehla throw 3 se kam hai, yeh diya hone par head aane ki probability nikaalo.
Solution dikhao
Answer $\frac{1}{4}$
Maan lo $F$ hai pehla throw 3 se kam, isliye $P (F) = \frac{2}{6} = \frac{1}{3}$ .
Head ke liye pehle throw par 1 chahiye (odd, isliye coin uchhala jaata hai) aur phir head: $P (E \cap F) = \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}$ . Pehle throw par 2 aane se die phenka jaata hai, coin kabhi nahi.
$P (E ∣ F) = \frac{1 / 12}{1 / 3} = \frac{1}{4}$ .
Quick answerEasyNCERT Exercise 13.1, Q16
$P (A) = 0.7$ hai aur $B$ ek impossible event hai. Tab $P (A ∣ B)$ hai
$0$ $0.7$ $1$ defined nahi hai
Answer dekho
Answer (D) defined nahi hai
$P (B) = 0$ hai, aur definition ko $P (B) \neq 0$ chahiye; fraction $\frac{0}{0}$ ho jaata. Isliye $P (A ∣ B)$ defined nahi hai.
Quick answerMediumNCERT Exercise 13.1, Q17
$A$ aur $B$ aise events hain jinke liye $P (A ∣ B) = P (B ∣ A) \neq 0$ . Inmein se kya zaroor sach hona chahiye?
$A \subset B$ $A \cap B = \emptyset$ $P (A) = P (B)$ $A \cup B = S$
Answer dekho
Answer (C) $P (A) = P (B)$
Dono fractions ka numerator $P (A \cap B)$ same hai, jo yahan zero nahi hai. Same non-zero numerator wale barabar fractions ke denominators bhi barabar hote hain, isliye $P (A) = P (B)$ . Events ka barabar ya ek doosre ke andar hona zaroori nahi, aur disjoint events dono sides ko 0 bana dete.

13.3 Probability par Multiplication Theorem

Aksar hum yeh chance chahte hain ki do cheezein dono hon: packet se nikali pehli do toffees orange hon, ya train late ho aur baarish bhi ho rahi ho. Event $E$ aur $F$ dono hai $E \cap F$ , jise kabhi-kabhi $E F$ bhi likhte hain. Conditional probability ki definition ko ulta karke hum ise ek-ek stage karke nikaal sakte hain.

(13.1) se, jab bhi $P (F) \neq 0$ ho, $P (E \cap F) = P (F) P (E ∣ F)$ . Dono events ke roles badal do aur $F \cap E = E \cap F$ use karo: yahi probability $P (E)$ guna $P (F ∣ E)$ bhi hai, bashart $P (E) \neq 0$ ho.

Multiplication rule

\begin{aligned} P (E \cap F) \\ = P (E) P (F ∣ E) \\ = P (F) P (E ∣ F) \end{aligned}

(13.2)

jahan

$P (F ∣ E)$: $E$ ho jaane ke baad $F$ ka chance

Yeh probability ka multiplication rule hai. Tree diagram par iska matlab hai path ke saath multiply karo: pehla factor pehli branch ka chance hai, aur doosra factor pehli branch diye hone par agli branch ka chance (Fig. 13.4).

3/3Dono orange: us path ke saath multiply karo, kul $\frac{4}{12} \cdot \frac{3}{11} = \frac{1}{11}$ .

Fig. 13.4Without replacement mein doosri branch ki probabilities pehle draw par depend karti hain; path ke saath multiply karo.NCERT Example 8 · p. 415

8 mango aur 4 orange toffees wale packet se do toffees nikaalne ka tree. Pehle step mein sirf pehla draw dikhta hai: mango 12 mein se 8 aur orange 12 mein se 4. Doosre step mein doosra draw dikhta hai: mango ke baad, mango 11 mein se 7 aur orange 11 mein se 4; orange ke baad, mango 11 mein se 8 aur orange 11 mein se 3. Teesre step mein path orange phir orange mota dikhaya gaya hai, baaki branches fade hoti hain, aur chaaron path products dikhte hain: 14 over 33, 8 over 33, 8 over 33 aur 4 over 12 times 3 over 11, jo 1 over 11 hai.

Example 8NCERT Example 8 · p. 415

Do orange toffees

Ek packet mein 8 mango aur 4 orange toffees hain. Ek ke baad ek do toffees nikaali jaati hain, aur pehli wapas nahi daali jaati. Dono ke orange hone ki probability kya hai?

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

Teen events ke liye hum rule do baar lagaate hain. Har naya factor un sab baaton par conditioned hota hai jo pehle ho chuki hain, aur yahi pattern chaar ya zyada events ke liye bhi chalta hai.

Teen events

\begin{aligned} P (E \cap F \cap G) \\ = P (E) P (F ∣ E) \\ \times P (G ∣ E \cap F) \end{aligned}

(13.3)

Example 9NCERT Example 9 · p. 416

Ek raffle draw

School ke raffle box ke 20 tickets mein se chhah par prize hai. Ek ke baad ek teen tickets without replacement nikaale jaate hain. Pehle do tickets jeetne aur teesra na jeetne ki probability nikaalo.

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

Quick check

3 mein se 1

Ek box mein 3 red aur 4 blue pens hain. Ek ke baad ek do pens nikaale jaate hain, aur pehla wapas nahi rakha jaata. Dono ke red hone ka chance nikaalo.

Answer dekho

Answer 0.142857

\frac{3}{7} \cdot \frac{2}{6} = \frac{1}{7} \approx 0.143

\frac{3}{7} \cdot \frac{3}{7}

lene ka matlab hai pehle pen ko aise maanna jaise woh wapas rakh diya gaya ho.

13.4 Independent Events

1 se 20 tak numbered box se random se ek ticket nikaalo. Maan lo $E$ hai number even hai aur $F$ hai number 5 ka multiple hai. Tab $P (E) = \frac{1}{2}$ . Agar tumhe bataya jaaye ki $F$ hua hai, to ticket 5, 10, 15 aur 20 mein se ek hai, aur inmein se do even hain, isliye $P (E ∣ F) = \frac{2}{4} = \frac{1}{2}$ . $F$ ki khabar ne $E$ ka chance nahi badla. Ulti taraf se bhi yahi hota hai: $P (F) = \frac{4}{20} = \frac{1}{5}$ aur $P (F ∣ E) = \frac{2}{10} = \frac{1}{5}$ .

Ab 1 se 15 tak numbered box lo, wahi do descriptions ke saath. Is baar $P (E) = \frac{7}{15}$ , lekin 5, 10 aur 15 mein sirf 10 even hai, isliye $P (E ∣ F) = \frac{1}{3}$ . Dono boxes mein events bilkul ek jaise lagte hain, phir bhi sirf pehla pair ek doosre se unaffected hai. Independence probabilities ke baare mein ek fact hai, is baare mein nahi ki do events judi hui lagti hain ya nahi.

Definition

Independent events

Events $E$ aur $F$ independent hain agar $P (F ∣ E) = P (F)$ (jab $P (E) \neq 0$ ) aur $P (E ∣ F) = P (E)$ (jab $P (F) \neq 0$ ): ek ke hone ka pata chalne se doosre ki probability nahi badalti.

ExampleTickets 1–20: even aur 5 ka multiple.

Multiplication rule se $P (E \cap F) = P (E) P (F ∣ E)$ , aur independent events ke liye doosra factor seedha $P (F)$ hai. Jo product form milta hai use $P (E)$ ya $P (F)$ par koi condition nahi chahiye, isliye yahi working definition hai.

Independent events

\begin{aligned} P (E \cap F) \\ = P (E) P (F) \end{aligned}

(13.4)

Jo events is test mein fail hote hain unhe dependent kehte hain: $P (E \cap F) \neq P (E) P (F)$ . Do experiments tab independent kehlaate hain jab pehle ka har event doosre ke har event se independent ho, jaise coin toss aur die throw, ya ek hi die ke do alag throws.

Ek area picture madad karti hai (Fig. 13.5). Sample space ko ek unit square ki tarah banao jismein area probability ko dikhata hai, aur $E$ ko uska left half maano. $F$ , $E$ se theek tab independent hai jab $F$ , $E$ ka utna hi fraction cover kare jitna poore square ka.

2/4 $F$ , $E$ ko proportion mein kaatta hai: overlap $\frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10}$ , product hi hai.

Fig. 13.5Independent events ek doosre ko proportion mein kaatte hain; exclusive events bilkul nahi milte.NCERT Fig. 13.3 · p. 420

Ek unit square sample space ko dikhata hai, jismein area probability ke barabar hai. Pehle step mein left half event E ki tarah shade hai, probability one half ke saath. Doosre step mein neeche ki ek band, height ka one fifth, event F hai; yeh E ko kaatti hai, aur overlap hatched hai, area one tenth, jo P of E times P of F ke barabar hai, isliye panel par independent likha hai. Teesre step mein F ka complement, upar ke four fifths, outline hota hai, aur E ke saath uska overlap two fifths area ka hai, phir se product, isliye E aur F prime bhi independent hain. Chauthe step mein F ko same area ke ek box ki tarah poori tarah right half ke andar banaya gaya hai, isliye woh E se nahi milta: overlap 0 hai jabki product one tenth hai, aur panel par exclusive likha hai, isliye dependent.

Compare

Independent ya mutually exclusive?

Independent

Probabilities se defined: $P (E \cap F) = P (E) P (F)$ .
Outcomes share kar sakte hain; positive probabilities ke saath karne hi padte hain.
Ek ko jaanne se doosre ke baare mein kuch naya pata nahi chalta.

Mutually exclusive

Sets se defined: $E \cap F = \emptyset$ .
Koi bhi outcome share nahi karte.
Ek ke hone ka matlab hai ki doosra nahi hua.

Isliye non-zero probabilities wale do events kabhi dono nahi ho sakte. Agar $E \cap F = \emptyset$ ho to $P (E \cap F) = 0$ , jabki $P (E) P (F) > 0$ . Is tarah ke mutually exclusive events hamesha dependent hote hain.

Example 10NCERT Example 10 · p. 419

Ek die, do descriptions

Ek die phenka jaata hai. Maan lo $A$ hai number zyada se zyada 2 hai aur $B$ hai number odd hai. Kya $A$ aur $B$ independent hain? Agar $A$ ki jagah zyada se zyada 3 ho to kya badalta hai?

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

Example 11NCERT Example 11 · p. 419

Pehle six, aur 7 ka total

Do dice phenke jaate hain. Maan lo $A$ hai pehle die par 6 aur $B$ hai total 7 hai. Check karo ki $A$ aur $B$ independent hain ya nahi.

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

Example 12NCERT Example 12 · p. 419

Kaunse pairs independent hain?

Do dice phenke jaate hain. $E$ : ek double (dono dice barabar). $F$ : total 8 hai. $G$ : pehla die even hai. Pairs $(E, F)$ , $(E, G)$ , $(F, G)$ mein se kaunse independent hain?

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

Example 13NCERT Example 13 · p. 420

Independent events ke complements

Dikhao ki complement lene par bhi independence bani rehti hai: jab bhi $E$ aur $F$ independent hon, to $E$ aur $F^{'}$ bhi independent hote hain.

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

Example 14NCERT Example 14 · p. 421

Do mein se kam se kam ek

Dikhao ki agar $A$ aur $B$ independent hain, to unmein se kam se kam ek hone ki probability $1 - P (A^{'}) P (B^{'})$ hai. Ek jewellery shop mein do alarms hain jo independently kaam karte hain; chori ke waqt woh 0.1 aur 0.2 probability se chup rehte hain. Kam se kam ek ke bajne ki probability nikaalo.

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

13.5 Bayes' Theorem

Teacher ki desk par pens ke do boxes rakhe hain. Box I mein 4 blue aur 2 black pens hain; Box II mein 3 blue aur 5 black. Ek student random se ek box chunta hai aur bina dekhe ek pen nikaalta hai. Seedha sawaal aasaan hai: agar pen Box II se aaya, to woh $\frac{5}{8}$ probability se black hai. Ab ise ulta karo. Tumhe sirf itna dikhta hai ki pen black hai. Iske Box II se aane ki kitni sambhavna hai? Hum ulta reasoning kar rahe hain, dikhe hue effect se uske possible cause tak.

English minister aur mathematician Thomas Bayes ne aise inverse sawaalon ka jawab dene ka tareeka dhoondha. Unka essay unki death ke do saal baad, 1763 mein, unke dost Richard Price ne Royal Society mein padha. Pehle do ideas aate hain: sample space ka partition, aur theorem of total probability.

13.5.1 Sample space ka partition

Definition

Sample space ka partition

Hum kehte hain ki events $E_{1}, E_{2}, \dots, E_{n}$ , $S$ ka partition karte hain jab (a) koi do overlap na karein, $i \neq j$ ke liye $E_{i} \cap E_{j} = \emptyset$ ; (b) unka union poora $S$ ho; aur (c) har ek ki probability positive ho, $P (E_{i}) > 0$ .

ExampleEk die ke liye: ${1, 2}$ , ${3, 4, 5}$ , ${6}$ .

Outcome kuch bhi ho, partition ka theek ek part hota hai. Pen wali problem mein Box I chuna gaya aur Box II chuna gaya do parts hain; Example 19 ke teen suppliers teen parts wala partition dete hain. $0 < P (E) < 1$ wala koi bhi event $E$ , $E^{'}$ ke saath, bhi ek partition hai, aur do events $E$ aur $F$ ek aur bareek partition dete hain: chaar tukdon $E \cap F$ , $E \cap F^{'}$ , $E^{'} \cap F$ , $E^{'} \cap F^{'}$ mein se jinki probability positive hai. Kyunki ek sample space ko kai tarah se kaata ja sakta hai, hum woh partition chunte hain jiske parts possible causes hon.

Fig. 13.6 teen events wale partition ko vertical strips ki tarah dikhata hai jinki widths $P (E_{1})$ , $P (E_{2})$ aur $P (E_{3})$ hain. Phir ek aur event $A$ har strip se ek tukda kaatta hai, aur strip $E_{i}$ mein us tukde ki height $P (A ∣ E_{i})$ hai.

3/4 $A$ ke tukde jodne par, $0.15 + 0.18 + 0.04$ , kul $P (A) = 0.37$ milta hai.

Fig. 13.6Partition strips ki tarah: total probability A ke tukde jodti hai; Bayes' theorem ek tukde ka share leti hai.NCERT Fig. 13.4 · p. 424

Ek rectangle sample space ko dikhata hai, jismein area probability ke barabar hai. Pehle step mein yeh teen vertical strips E1, E2 aur E3 mein bata hai jinki widths 0.5, 0.3 aur 0.2 hain, rectangle ke neeche label ki hui. Doosre step mein har strip ke neeche event A ka ek hatched tukda dikhta hai, heights 0.3, 0.6 aur 0.2 ke saath, jo har strip mein A ki conditional probabilities hain. Teesre step mein tukde apne areas 0.15, 0.18 aur 0.04 dikhate hain, aur label par likha hai P of A equals 0.37. Chauthe step mein sirf beech wala tukda bright rehta hai aur label par likha hai P of E2 given A equals 0.18 over 0.37, lagbhag 0.49.

13.5.2 Total probability ka theorem

TheoremTotal probability ka theorem

Maan lo events $E_{1}, E_{2}, \dots, E_{n}$ , $S$ ka partition karte hain, aur $A$ koi bhi event hai. Tab $P (A)$ , $n$ products $P (E_{j}) P (A ∣ E_{j})$ ke sum ke barabar hai, har part ke liye ek.

$E_{j}$ milkar $S$ ko cover karte hain, isliye $A = A \cap S$ tukdon $A \cap E_{1}$ , $A \cap E_{2}$ , …, $A \cap E_{n}$ ka union hai. Har tukda $A \cap E_{j}$ , $E_{j}$ ke andar baitha hai, aur $E_{j}$ disjoint hain, isliye tukde bhi disjoint hain. Isliye unki probabilities jud jaati hain: $P (A)$ , $n$ numbers $P (A \cap E_{j})$ ka sum hai.

Kyunki har $P (E_{j}) > 0$ hai, multiplication rule se $P (A \cap E_{j}) = P (E_{j}) P (A ∣ E_{j})$ milta hai. Substitute karne se proof poora ho jaata hai.

Total probability

\begin{aligned} P (A) \\ = \sum_{j = 1}^{n} P (E_{j}) \\ \times P (A ∣ E_{j}) \end{aligned}

(13.5)

Strip picture mein $A$ ka har tukda $P (E_{j})$ width aur $P (A ∣ E_{j})$ height ka rectangle hai, isliye theorem areas jodta hai. Tree diagram par yeh un sab paths ki probabilities jodta hai jo $A$ par khatam hote hain (Fig. 13.7).

3/3Dono raaste on time khatam ho sakte hain, isliye dono paths jodo: kul $0.12 + 0.595 = 0.715$ .

Fig. 13.7Tree par total probability: event par khatam hone wale har path ko jodo.NCERT p. 425

Do pehli branches wala tree: fog 0.3 ke saath aur clear 0.7 ke saath. Fog ke baad train 0.4 se on time aur 0.6 se late hai; clear subah ke baad 0.85 se on time aur 0.15 se late. Doosre step mein leaves ke paas path probabilities dikhti hain: 0.12, 0.18, 0.595 aur 0.105. Teesre step mein dono on-time paths highlight hote hain, late paths fade hote hain, aur label par likha hai sum 0.715.

Example 15NCERT Example 15 · p. 425

Fog aur subah ki train

Delhi mein December ki subah 0.3 probability se ghana fog hota hai. Ek train foggy subah 0.4 probability se aur clear subah 0.85 probability se time par pahunchti hai. Train ke time par pahunchne ki probability nikaalo.

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

Bayes' theorem

TheoremBayes' theorem

Maan lo events $E_{1}, E_{2}, \dots, E_{n}$ , $S$ ka partition karte hain aur $A$ ek event hai jiske liye $P (A) \neq 0$ . Tab har $i$ ke liye $P (E_{i} ∣ A)$ barabar hai $P (E_{i}) P (A ∣ E_{i})$ ko $\sum_{j = 1}^{n} P (E_{j}) P (A ∣ E_{j})$ se divide karne ke.

Bayes' theorem

\begin{aligned} P (E_{i} ∣ A) \\ = \frac{P (E_{i}) P (A ∣ E_{i})}{P (A)} \end{aligned}

(13.6)

jahan

$P (A)$: evidence ki total probability, $\sum_{j} P (E_{j}) P (A ∣ E_{j})$ , (13.5) se

Bayes' theorem ko ek share ki tarah padho. Numerator $E_{i}$ se hokar jaane wale us ek path ki probability hai jo $A$ par khatam hota hai; denominator un sab paths ki probability hai jo $A$ par khatam hote hain. Fig. 13.6 ke aakhri step mein yeh $A$ ke ek tukde ka area hai, poore $A$ ke area se divide kiya hua.

Example 16NCERT Example 16 · p. 426

Pen kis box se aaya?

Box I mein 4 blue aur 2 black pens hain; Box II mein 3 blue aur 5 black. Random se ek box chuna jaata hai aur usmein se nikala pen black hai. Uske Box II se aane ki probability nikaalo.

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

Example 17NCERT Example 17 · p. 426

Coins ki teen pouches

Teen ek jaisi pouches mein se har ek mein teen coins hain. Pouch I mein 3 gold coins hain, Pouch II mein 1 gold aur 2 silver, aur Pouch III mein 3 silver. Random se ek pouch chuni jaati hai aur usmein se nikala ek coin gold hai. Us pouch mein bache dono coins ke gold hone ki probability kya hai?

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

Example 18NCERT Example 18 · p. 427

Ek positive screening test

Ek screening test aisi condition ke liye use hota hai jo population ke 1% logon ko hoti hai. Agar kisi ko condition hai, to test 0.98 probability se positive aata hai; agar nahi hai, to bhi 0.03 probability se false positive deta hai. Random se chuna gaya ek vyakti positive test karta hai. Us vyakti ko condition hone ki probability kya hai?

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

3/3395 positive tests mein se sirf 98 beemar logon ke hain: lagbhag 0.25.

Fig. 13.8Natural frequencies Bayes' theorem ko saaf dikhati hain: yahan zyadatar positives bade healthy group se aate hain.NCERT Example 18 · p. 427

Counts ka ek tree. Sabse upar 10 000 logon ka ek box hai. Woh 100 beemar aur 9 900 theek mein bantta hai. Doosre step mein har group test result se bantta hai: 100 beemar 98 positive aur 2 negative tests dete hain; 9 900 theek log 297 positive aur 9 603 negative tests dete hain. Teesre step mein negative boxes fade hote hain aur neeche ek bar saare 395 positive tests dikhata hai, 98 beemar logon ka chhota hissa aur 297 theek logon ka lamba hissa.

Example 18 wala surprise medicine, airport security aur spam filters mein aam hai. Jab koi condition rare ho, to achha test bhi beemar logon se kahin zyada healthy logon se milta hai, isliye uske thode se false alarms uske true alarms se zyada ho sakte hain. Neeche wala lab tumhe teeno numbers badalne aur posterior ka response dekhne deta hai.

InteractiveBayes lab

Screening test ke liye ek interactive Bayes calculator. Teen sliders set karte hain beemar hone ki prior probability (log scale par 0.1% se 99.9%), beemar hone par positive test ka chance, aur theek hone par positive test ka chance. Preset buttons Example 18, ek bahut rare condition aur ek common condition load karte hain. 10 000 logon ke counts ka ek tree dikhata hai ki kitne beemar aur kitne theek hain aur har group mein kitne positive ya negative aate hain, aur ek bar positive tests ko beemar aur theek logon mein baantta hai. Ek status panel positive test ki total probability, positive test diye hone par beemar hone ki posterior probability, aur negative test diye hone par theek hone ki probability deta hai. Test again button prior ko current posterior se badal deta hai. Jab tak learner kuch na kare, kuch nahi hilta.

Fig. 13.9Teeno sliders drag karo aur posterior dekho. Test again answer ko naye prior ki tarah use karta hai, yeh maankar ki true state diye hone par doosra test pehle se independent hai.

Example 19NCERT Example 19 · p. 428

Kharab battery kis supplier ne banayi?

Ek phone factory teen suppliers se batteries khareedti hai: 45% pehle se, 35% doosre se aur 20% teesre se. Unki batteries 2%, 3% aur 4% baar faulty hoti hain. Store se random se uthayi gayi ek battery faulty hai. Uske doosre supplier se aane ki probability nikaalo.

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

Example 20NCERT Example 20 · p. 429

School ke liye late

Tanvi aadhe dinon metro se school jaati hai, 30% dinon auto-rickshaw se aur baaki dinon cycle se. Metro se woh $\frac{1}{10}$ probability se late hoti hai, auto se $\frac{1}{4}$ aur cycle se $\frac{1}{20}$ . Ek din woh late pahunchti hai. Uske auto se aane ki probability kya hai?

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

Example 21NCERT Example 21 · p. 430

Ek report jo jhoothi ho sakti hai

Ravi 5 mein se 4 baar sach bolta hai. Woh ek die phenkta hai aur tumhe batata hai ki number 4 se bada tha. Sach mein aisa hone ki probability nikaalo. (Jab Ravi jhooth bolta hai, to woh doosri kism ka result batata hai.)

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

Quick check

4 mein se 1

Box I mein 1 red aur 3 white balls hain; Box II mein 3 red aur 1 white. Random se ek box uthaya jaata hai aur usmein se nikali ball red hai. Uske Box II se aane ka chance nikaalo.

Answer dekho

Answer 0.75

\frac{\frac{1}{2} \cdot \frac{3}{4}}{\frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}} = \frac{3}{4}

. Yahan forward probability

P (red ∣ Box II)

bhi

\frac{3}{4}

hai, lekin sirf isliye ki dono boxes ek doosre ki mirror image hain.

Random variables

Bahut se experiments ek aise number par khatam hote hain jo humein raw outcome se zyada important hai: do dice ka total, ek carton mein faulty items ki sankhya, fair stall par jeete paise. Ek coin teen baar uchhaalo. Sample space mein 8 outcomes hain, aur har ek ke saath hum usmein aaye heads ki sankhya jod sakte hain.

Definition

Random variable

Ek rule jo random experiment ke har outcome ke saath ek real number jodta hai; doosre shabdon mein, sample space $S$ se real numbers $𝐑$ tak ek function.

ExampleTeen tosses: $X$ = heads ki sankhya, isliye $X (H H T) = 2$ .

Teen tosses ke liye $X (T T T) = 0$ , $X (T H T) = 1$ , $X (H H T) = 2$ aur $X (H H H) = 3$ (Fig. 13.10). Ek sample space par bahut se random variables ho sakte hain. Maan lo ek game har head ke ₹10 deta hai aur har tail ke ₹5 leta hai. Tumhari jeet $Y = 10 X - 5 (3 - X)$ hai, yaani $Y = 15 X - 15$ rupaye. Isliye $Y (T T T) = - 15$ , $Y (H T T) = 0$ , $Y (H H T) = 15$ aur $Y (H H H) = 30$ . $X$ aur $Y$ ek hi $S$ par alag functions hain.

2/3 $X$ har outcome ko uske heads ki sankhya par bhejta hai.

Fig. 13.10Random variable har outcome ko ek number par bhejta hai; unhi outcomes par doosra rule doosra random variable deta hai.NCERT p. 431

Upar, teen tosses ke aath outcomes S label wale shaded box mein hain: TTT, HTT, THT, TTH, HHT, HTH, THH aur HHH. Doosre step mein har outcome se neeche ek number line tak arrow jaata hai jis par ticks 0, 1, 2 aur 3 hain: TTT se 0, ek head wale teen outcomes se 1, do heads wale teen se 2, aur HHH se 3, note count the heads ke saath. Teesre step mein wahi arrows rehte hain lekin tick labels badal kar minus 15, 0, 15 aur 30 ho jaate hain, note 10 rupees per head, minus 5 rupees per tail ke saath.

Aage ki jhalak

Aage ki jhalak: probability distribution

$X$ ki values ko unki probabilities ke saath list karo, to tumhare paas $X$ ka probability distribution hai:

Table 13.1Teen tosses mein heads ki sankhya

$x$	$P (X = x)$
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

Probabilities ka total 1 hai, kyunki har outcome theek ek value deta hai. Current syllabus random variable ki definition par ruk jaata hai; distributions, unka mean aur variance aage ke courses mein padhe jaate hain.

Quick check

2 mein se 1

Do dice phenke jaate hain aur

X

dikhe hue do numbers mein se bada number hai.

X ((2, 5))

kya hai?

2573

Answer dekho

Answer (B) 5

X

bada number chunta hai, 5. Total 7 aur difference 3 usi sample space par doosre random variables honge.

Practice · 14 questions

Practice 13.3: answers ke saath

1
Short answerHardNCERT Exercise 13.3, Q1
Ek jar mein 4 red aur 6 green balls hain. Random se ek ball uthao, uska colour note karo aur use usi colour ki 3 extra balls ke saath jar mein wapas daal do. Phir doosri ball uthao.
1. Doosri ball ke green hone ki probability nikaalo.
  Answer dekho
  Answer 0.6
  $P (G_{2}) = \frac{6}{10} \cdot \frac{9}{13} + \frac{4}{10} \cdot \frac{6}{13} = \frac{54 + 24}{130} = \frac{3}{5}$ , jo pehli ball ke barabar hi hai.
2. Doosri ball green hai, yeh diya hone par pehli ball ke green hone ki probability nikaalo.
  Answer dekho
  Answer 0.69231
  $P (G_{1} ∣ G_{2}) = \frac{(6 / 10) (9 / 13)}{3 / 5} = \frac{9}{13}$ .
Short answerEasyNCERT Exercise 13.3, Q2
Basket I mein 5 pake aur 3 kachche mangoes hain; Basket II mein 2 pake aur 6 kachche. Random se ek basket chuni jaati hai aur usmein se nikala mango paka hai. Uske Basket I se aane ki probability nikaalo.
Solution dikhao
Answer $\frac{5}{7}$
$P (ripe ∣ I) = \frac{5}{8}$ aur $P (ripe ∣ I I) = \frac{2}{8}$ .
Barabar priors cancel ho jaate hain, isliye Basket I ka posterior $\frac{5 / 8}{5 / 8 + 2 / 8} = \frac{5}{7}$ hai.
Short answerMediumNCERT Exercise 13.3, Q3
Ek coaching centre mein 70% students online padhte hain aur 30% in person aate hain. Ek mock test mein online students mein se 25% aur in-person students mein se 40% ke 90 se zyada marks aate hain. 90 se zyada score karne wala ek student random se chuna jaata hai. Student ke in person aane ki probability nikaalo.
Solution dikhao
Answer $\frac{24}{59} \approx 0.41$
90 se zyada tak ke paths: online $0.7 \times 0.25 = 0.175$ ; in person $0.3 \times 0.4 = 0.12$ .
Dono paths ka total $0.295$ hai, isliye in person ka posterior $\frac{0.12}{0.295} = \frac{24}{59}$ hai.
Short answerMediumNCERT Exercise 13.3, Q4
Ek test mein har multiple-choice question ke 5 options hain. Ek student $\frac{2}{3}$ probability se answer jaanta hai aur warna random se guess karta hai. Yeh diya hai ki student ne question sahi kiya; student ke answer jaanne ki probability kya hai?
Solution dikhao
Answer $\frac{10}{11}$
$P (correct) = \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{5} = \frac{11}{15}$ .
Jaanta tha ka posterior $\frac{2 / 3}{11 / 15} = \frac{10}{11}$ hai.
Short answerHardNCERT Exercise 13.3, Q5
Ek airport scanner un bags mein se 95% ko flag karta hai jinmein banned item hai, aur un bags mein se bhi 2% ko flag karta hai jinmein nahi hai. 200 mein se ek bag mein banned item hai. Ek bag flag hota hai. Usmein sach mein banned item hone ki probability kya hai?
Solution dikhao
Answer $\frac{95}{493} \approx 0.19$
True alarms $0.005 \times 0.95 = 0.00475$ hain. False alarms $0.995 \times 0.02 = 0.0199$ hain.
Saare flagged bags: $0.00475 + 0.0199 = 0.02465$ . Isliye banned item ka chance $\frac{0.00475}{0.02465} = \frac{95}{493}$ hai, lagbhag 0.19.
Yaani lagbhag paanch mein se chaar flagged bags false alarms hain, isiliye flagged bag ko phir haath se check kiya jaata hai.
Short answerMediumNCERT Exercise 13.3, Q6
Ek bag mein teen dice hain: ek ordinary die, ek die jiske teen faces par 6 hai, aur ek die jiske har face par 6 hai. Random se ek die chuna jaata hai aur phenka jaata hai, aur us par 6 aata hai. Uske har face par 6 wala die hone ki probability kya hai?
Solution dikhao
Answer $\frac{3}{5}$
Har ek ka prior $\frac{1}{3}$ ho to $P (6) = \frac{1}{3} (\frac{1}{6} + \frac{1}{2} + 1) = \frac{5}{9}$ .
All-six die ka posterior $\frac{1 / 3}{5 / 9} = \frac{3}{5}$ hai.
Short answerMediumNCERT Exercise 13.3, Q7
Ek delivery company mein 300 cyclists, 500 motorbike riders aur 200 van drivers kaam karte hain. Ek mahine mein traffic fine ki probability cyclist ke liye 0.02, rider ke liye 0.05 aur van driver ke liye 0.1 hai. Ek employee ko is mahine fine mila. Employee ke cyclist hone ki probability nikaalo.
Solution dikhao
Answer $\frac{2}{17}$
Expected fines: cyclists $300 \times 0.02 = 6$ , riders $500 \times 0.05 = 25$ , vans $200 \times 0.1 = 20$ ; kul 51.
Isliye cyclist ka chance $\frac{6}{51} = \frac{2}{17}$ hai.
Quick answerMediumNCERT Exercise 13.3, Q8
Ek office mein do printers hain. Printer P 70% pages print karta hai aur printer Q baaki. P ka page 0.03 probability se smudged hota hai, aur Q ka 0.05 probability se. Out-tray se uthaya gaya ek page smudged hai. Use Q ne print kiya ho, iski probability kya hai?
Answer dekho
Answer 0.41667
Paths: $0.7 \times 0.03 = 0.021$ aur $0.3 \times 0.05 = 0.015$ . Isliye $P (Q ∣ smudged) = \frac{0.015}{0.036} = \frac{5}{12}$ .
Quick answerMediumNCERT Exercise 13.3, Q9
Do candidates, X aur Y, school captain ke liye khade hain; X 0.55 probability se jeetta hai aur Y 0.45 se. Agar X jeete, to 0.8 probability se ek naya sports period shuru hota hai; agar Y jeete, to 0.4 probability se. Sports period shuru hua. Y ke jeetne ki probability kya hai?
Answer dekho
Answer 0.29032
Paths: $0.55 \times 0.8 = 0.44$ aur $0.45 \times 0.4 = 0.18$ . Isliye $P (Y ∣ period) = \frac{0.18}{0.62} = \frac{9}{31}$ .
Short answerHardNCERT Exercise 13.3, Q10
Arjun ek die phenkta hai. Agar 1 ya 2 aaye, to woh coin teen baar uchhalta hai; warna do baar. Woh tumhe batata hai ki use theek do heads mile. Die par 1 ya 2 aane ki probability kya hai?
Solution dikhao
Answer $\frac{3}{7}$
Paths: 1 ya 2 se hokar $\frac{1}{3} \cdot \frac{3}{8} = \frac{1}{8}$ , aur 3 se 6 se hokar $\frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}$ .
Isliye 1 ya 2 ka posterior $\frac{1 / 8}{1 / 8 + 1 / 6} = \frac{3}{7}$ hai.
Short answerMediumNCERT Exercise 13.3, Q11
Ek bakery teen shifts chalati hai. Morning shift 40% loaves banati hai, afternoon shift 35% aur night shift 25%. Jale hue loaves ka proportion 2%, 3% aur 6% hai. Ek jala hua loaf milta hai. Use night shift ne banaya ho, iski probability kya hai?
Solution dikhao
Answer $\frac{30}{67} \approx 0.45$
Burnt tak ke paths: $0.4 \times 0.02 = 0.008$ , $0.35 \times 0.03 = 0.0105$ , $0.25 \times 0.06 = 0.015$ ; total $0.0335$ .
Isliye night shift ka posterior $\frac{0.015}{0.0335} = \frac{30}{67}$ hai.
Short answerHardNCERT Exercise 13.3, Q12
52 cards ke poore pack se ek card kho jaata hai. Phir bache 51 mein se do cards nikaale jaate hain aur dono red hain. Khoye card ke red hone ki probability nikaalo.
Solution dikhao
Answer $\frac{12}{25}$
Red card khoya, yeh diya hone par: $P (two red) = \frac{25}{51} \cdot \frac{24}{50}$ . Black card khoya, yeh diya hone par: $\frac{26}{51} \cdot \frac{25}{50}$ .
Priors $\frac{1}{2}$ aur common factor $\frac{1}{51 \cdot 50}$ cancel ho jaate hain, bachta hai $\frac{25 \cdot 24}{25 \cdot 24 + 26 \cdot 25} = \frac{600}{1250}$ , jo $\frac{12}{25}$ hai.
Draw mein do red cards aane se red card ke gayab hone ki sambhavna thodi kam ho jaati hai.
Quick answerEasyNCERT Exercise 13.3, Q13
Meena 3 mein se 2 baar sach bolti hai. Ek coin uchhala jaata hai aur woh tail batati hai. Sach mein tail hone ki probability hai
$\frac{2}{3}$ $\frac{1}{3}$ $\frac{1}{2}$ $\frac{5}{6}$
Answer dekho
Answer (A) $\frac{2}{3}$
$\frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{1}{2} \cdot \frac{2}{3} + \frac{1}{2} \cdot \frac{1}{3}} = \frac{2}{3}$ . Fair coin ke saath posterior uske sach bolne ke rate ke barabar hai.
Quick answerMediumNCERT Exercise 13.3, Q14
$B \subset A$ aur $P (A) \neq 0$ hai. Kaunsa statement sahi hai?
$P (B ∣ A) < P (B)$ hamesha $P (B ∣ A) = P (B)$ $P (B ∣ A) \geq P (B)$ $P (A ∣ B) < 1$
Answer dekho
Answer (C) $P (B ∣ A) \geq P (B)$
$P (B ∣ A) = \frac{P (B)}{P (A)} \geq P (B)$ kyunki $P (A) \leq 1$ . Saath hi jab bhi $P (B) \neq 0$ ho, $P (A ∣ B) = 1$ , isliye aakhri option galat hai.

Miscellaneous Examples

Example 22NCERT Example 22 · p. 433

Sweets ke chaar jars

Counter par sweets ke chaar jars rakhe hain. Unmein kya hai, woh neeche list kiya gaya hai. Random se ek jar chuna jaata hai aur usmein se bina dekhe ek sweet nikaali jaati hai. Woh laddoo hai. Uske Jar II se aane ki probability nikaalo. Kaunsa jar sabse kam likely hai?

Table 13.2Har jar mein sweets

	Laddoo	Barfi	Peda
Jar I	2	3	5
Jar II	4	1	1
Jar III	1	4	3
Jar IV	3	3	6

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

Example 23NCERT Example 23 · p. 434

Mele mein ring toss

Mele ke ek ring-toss stall par Aman aur Bela baari-baari khelte hain, pehle Aman, jab tak unmein se koi ek bottle par ring na daal de aur jeet jaaye. Har throw par Aman $\frac{1}{3}$ probability se aur Bela $\frac{1}{4}$ probability se succeed karti hai, independently. Har player ke jeetne ki probability nikaalo.

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

3/3Saare tukde jodne par Aman kul $\frac{2}{3}$ probability se jeetta hai aur Bela $\frac{1}{3}$ se.

Fig. 13.11Har round jo bacha hai use usi ratio, 2 to 1, mein baantta hai, isliye Aman ka total two thirds hai.NCERT Example 23 · p. 434

Ek lamba bar total probability 1 dikhata hai. Pehle step mein uska pehla third A shaded hai, Aman ke pehle throw par jeetne ke liye, agla sixth dashed B hai, Bela ke pehle throw par jeetne ke liye, aur bacha hua half game goes on mark kiya gaya hai. Doosre step mein bacha half usi tarah bantta hai: A ko ek sixth aur B ko ek twelfth milta hai, ek quarter bachta hai. Teesre step mein aage ke rounds bar ko chhote se chhote A aur B tukdon se bharte hain, aur neeche ek doosra bar unhe ikattha karta hai: A two thirds jeetta hai aur B one third.

Example 24NCERT Example 24 · p. 435

Ek packet ke do seeds

Ek nursery ke seed packets 0.75 probability se achhe batch se aate hain. Achhe batch mein har seed 0.9 probability se ugta hai; kharab batch mein 0.5 probability se. Seeds ek doosre se independently ugte hain. Tum ek packet ke do seeds bote ho aur dono ug jaate hain. Packet ke achhe batch se hone ki probability nikaalo.

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

Quick check

2 mein se 1

Example 23 ke ring-toss game mein Bela ke apne bilkul pehle throw par jeetne ki probability kya hai?

Answer dekho

Answer 0.166667

Pehle Aman ko miss karna hoga (

\frac{2}{3}

) aur phir Bela ko succeed karna hoga (

\frac{1}{4}

\frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}

. Answer

\frac{1}{4}

yeh bhool jaata hai ki Aman pehle phenkta hai.

Timeline

Historical note: dice tables se axioms tak

c. 1564
Cardano dice ke outcomes ginte hain
Girolamo Cardano ki Liber de Ludo Aleae ne dice ke throws ke favourable cases list kiye; yeh sirf 1663 mein print hui.
c. 1620
Galileo aur teen dice
Galileo ne samjhaya ki teen dice ke saath total 10, total 9 se zyada baar kyun aata hai: 216 ordered outcomes mein se 27 banaam 25.
1654
Pascal aur Fermat
Blaise Pascal aur Pierre de Fermat ke beech ke letters ne yeh suljhaya ki adhure game ke stakes kaise baante jaayein.
1657
Pehli printed textbook
Christiaan Huygens ne De Ratiociniis in Ludo Aleae publish ki, games of chance par problems ki ek chhoti si book.
1713
Bernoulli ki Ars Conjectandi
Jacob Bernoulli ki book, jo unki death ke baad publish hui, ne law of large numbers ka ek shuruaati roop prove kiya.
1763
Bayes aur inverse probability
Effects se wapas causes tak reasoning par Thomas Bayes ka essay, Bayes ki death ke do saal baad, Richard Price ne present kiya.
1812
Laplace ka treatise
Pierre-Simon Laplace ne us rule ko general form mein likha jise aaj hum Bayes' theorem kehte hain.
1933
Kolmogorov ke axioms
Andrey Kolmogorov ne probability ko outcomes ke sets ke baare mein kuch axioms par khada kiya, yahi approach tumne Class 11 mein dekhi.

Recap

Summary aur key points

Yeh jaanna ki $F$ hua, $F$ ko naya sample space bana deta hai: $P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$ , jahan $P (F) \neq 0$ (Section 13.2).
$F$ diya hone par probabilities normal rules maanti hain: $P (S ∣ F) = 1$ , $P (E^{'} ∣ F) = 1 - P (E ∣ F)$ , aur addition rule bhi chalta hai (Section 13.2.1).
$P (E ∣ F)$ aur $P (F ∣ E)$ ka numerator same hai lekin denominators alag; aam taur par yeh alag hote hain.
Multiplication rule: $P (E \cap F) = P (E) P (F ∣ E)$ ; teen events ke liye stages mein multiply karo (Section 13.3).
Tree diagram path ke saath multiply karta hai aur paths ke across jodta hai (Fig. 13.3).
$E$ aur $F$ independent hain jab $P (E \cap F) = P (E) P (F)$ ; tab complements bhi independent hote hain (Section 13.4).
Positive probabilities wale mutually exclusive events dependent hote hain; teen events ke liye chaaron product checks chahiye.
Partition positive probabilities wale disjoint events ka aisa set hai jo milkar $S$ ko bhar de (Section 13.5.1).
Total probability ka formula: $P (A) = \sum_{j} P (E_{j}) P (A ∣ E_{j})$ (Section 13.5.2).
Bayes' theorem $P (A)$ mein ek path ka share lekar priors $P (E_{i})$ ko posteriors $P (E_{i} ∣ A)$ mein badalta hai (Bayes' theorem).
Rare condition ke saath zyadatar positive tests false ho sakte hain; check karne ke liye natural frequencies se gino (Example 18).
Random variable har outcome ke saath ek real number jodta hai; ek sample space par kai random variables ho sakte hain (Random variables).

Flash cards · 6

Formula cards

1 / 6

Chapter 13 par Miscellaneous Exercise

Practice · 17 questions

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

1
Short answerEasyNCERT Miscellaneous Exercise, Q1
$A$ aur $B$ aise events hain jinke liye $P (B) \neq 0$ . $P (A ∣ B)$ nikaalo agar
1. $B \subset A$ ;
  Solution dikhao
  Answer1
  $A \cap B = B$ , isliye $P (A ∣ B) = \frac{P (B)}{P (B)} = 1$ .
2. $A \cap B = \emptyset$ ;
  Solution dikhao
  Answer0
  $P (A \cap B) = 0$ , isliye $P (A ∣ B) = 0$ .
3. $A \subset B$ .
  Solution dikhao
  Answer $\frac{P (A)}{P (B)}$
  $A \cap B = A$ , isliye $P (A ∣ B) = \frac{P (A)}{P (B)}$ , jo kam se kam $P (A)$ hai.
2
Short answerMediumNCERT Miscellaneous Exercise, Q2
Do dice phenke jaate hain. Dono par 6 aane ki probability nikaalo, yeh diya hone par ki
1. kam se kam ek die par 6 hai;
  Answer dekho
  Answer 0.0909
  Kam se kam ek 6 ke 11 outcomes hain, aur sirf $(6, 6)$ mein do hain: $\frac{1}{11}$ .
2. pehle die par 6 hai.
  Answer dekho
  Answer 0.16667
  Pehla 6 diya hai, to doosre die par 6 aana chahiye: $\frac{1}{6}$ .
Short answerMediumNCERT Miscellaneous Exercise, Q3
Lagbhag 8% men aur 0.5% women red–green colour-blind hote hain. Barabar men aur women wale group se random se ek colour-blind vyakti chuna jaata hai. Us vyakti ke man hone ki probability kya hai?
Solution dikhao
Answer $\frac{16}{17}$
Barabar priors cancel ho jaate hain. Colour-blind paths mein men 0.08 aur women 0.005 dete hain, isliye man hone ka chance $\frac{0.08}{0.085} = \frac{16}{17}$ hai, lagbhag 0.94.
Short answerHardNCERT Miscellaneous Exercise, Q4
Packet ka har seed 0.8 probability se ugta hai, baaki seeds se independently. Chhah seeds boye jaate hain. Unmein se zyada se zyada 4 ke ugne ki probability nikaalo.
Solution dikhao
Answer $0.34464$
Saare 6 ugte hain: ${0.8}^{6} = 0.262144$ .
Theek 5 ugte hain: fail hone wala ek seed 6 tareekon se chuno, har ek ki probability ${0.8}^{5} \times 0.2$ , isliye $6 \times 0.065536 = 0.393216$ .
Zyada se zyada 4: $1 - 0.262144 - 0.393216 = 0.34464$ .
Quick answerMediumNCERT Miscellaneous Exercise, Q5
Random se ek leap year chuna jaata hai. Usmein 53 Fridays ya 53 Saturdays hone ki probability kya hai?
Answer dekho
Answer 0.42857
Do extra din 7 equally likely pairs (Sun–Mon, …, Sat–Sun) mein se ek hain. Thu–Fri, Fri–Sat aur Sat–Sun mein Friday ya Saturday hai: $\frac{3}{7}$ .
Short answerMediumNCERT Miscellaneous Exercise, Q6
Chaar boxes K, L, M aur N mein se har ek mein 10 marbles hain, table ke hisaab se rangon wale. Random se ek box chuna jaata hai aur usmein se nikala marble green hai. Uske box K, box L aur box M se aane ki probability nikaalo.
Table 13.3Har box mein marbles
Green Red White
K 2 5 3
L 4 4 2
M 7 1 2
N 0 3 7
Solution dikhao
Answer $\frac{2}{13}$ , $\frac{4}{13}$ , $\frac{7}{13}$
K, L, M aur N ke liye $P (green ∣ box)$ 0.2, 0.4, 0.7 aur 0 hai; priors $\frac{1}{4}$ cancel ho jaate hain.
Total $0.2 + 0.4 + 0.7 + 0 = 1.3$ , isliye posteriors $\frac{2}{13}$ , $\frac{4}{13}$ aur $\frac{7}{13}$ hain (aur N ke liye 0).
Short answerMediumNCERT Miscellaneous Exercise, Q7
Sardi mein ek student ko cold hone ka chance 40% hai. Daily walk is chance ko 0.8 se multiply karti hai; daily vitamin tablets ise 0.6 se multiply karti hain. Har student dono mein se theek ek chunta hai, barabar probability se. Ek student ko cold hua. Student ne walk chuni ho, iski probability nikaalo.
Solution dikhao
Answer $\frac{4}{7}$
Walk: $0.4 \times 0.8 = 0.32$ . Tablets: $0.4 \times 0.6 = 0.24$ .
Barabar priors $\frac{1}{2}$ cancel ho jaate hain, isliye student ke walk chunne ka chance $\frac{0.32}{0.32 + 0.24} = \frac{4}{7}$ hai.
Quick answerMediumNCERT Miscellaneous Exercise, Q8
$2 \times 2$ matrix ki har entry 0 ya 1 hai, independently, har ek $\frac{1}{2}$ probability se chuni gayi. Determinant ke non-zero hone ki probability kya hai?
Answer dekho
Answer 0.375
16 equally likely matrices hain. $a d - b c = 1$ ke liye $a = d = 1$ aur $b c = 0$ chahiye: 3 matrices. $a d - b c = - 1$ bhi isi tarah 3 deta hai. Isliye $\frac{6}{16} = \frac{3}{8}$ .
9
Short answerMediumNCERT Miscellaneous Exercise, Q9
Ek scooter start nahi hota agar uski battery (A) ya starter (B) fail ho jaaye. Records se: $P (A fails) = 0.25$ , $P (B fails but A does not) = 0.1$ , $P (both fail) = 0.05$ . Nikaalo
1. $P (A fails ∣ B has failed)$ ;
  Answer dekho
  Answer 0.33333
  $P (B fails) = 0.1 + 0.05 = 0.15$ , isliye $\frac{0.05}{0.15} = \frac{1}{3}$ .
2. $P (A fails but B does not)$ .
  Answer dekho
  Answer 0.2
  $0.25 - 0.05 = 0.2$ .
Short answerHardNCERT Miscellaneous Exercise, Q10
Bag I mein 4 white aur 2 black balls hain; Bag II mein 3 white aur 5 black. Bag I se random se uthayi gayi ek ball Bag II mein daal di jaati hai. Phir Bag II se ek ball nikaali jaati hai, aur woh white hai. Daali gayi ball ke white hone ki probability nikaalo.
Solution dikhao
Answer $\frac{8}{11}$
White daali ( $\frac{4}{6}$ ): tab Bag II mein 9 mein se 4 white. Black daali ( $\frac{2}{6}$ ): 9 mein se 3 white.
Paths: $\frac{4}{6} \cdot \frac{4}{9} = \frac{16}{54}$ aur $\frac{2}{6} \cdot \frac{3}{9} = \frac{6}{54}$ . Isliye white ball daalne ka chance $\frac{16}{22} = \frac{8}{11}$ hai.
Quick answerEasyNCERT Miscellaneous Exercise, Q11
$P (A) \neq 0$ aur $P (B ∣ A) = 0$ hai. Tab
$A \subset B$ $B \subset A$ $P (A \cap B) = 0$ $P (B) = 0$
Answer dekho
Answer (C) $P (A \cap B) = 0$
$P (A \cap B) = P (A) \times 0 = 0$ . $P (B) = 0$ hone ki koi majboori nahi: jab $A$ na ho, tab $B$ ab bhi ho sakta hai.
Quick answerMediumNCERT Miscellaneous Exercise, Q12
$P (A)$ aur $P (B)$ non-zero hain aur $P (A ∣ B) < P (A)$ . Kaunsa sahi hai?
$P (B ∣ A) > P (B)$ $P (B ∣ A) = P (B)$ $A$ aur $B$ independent hain $P (B ∣ A) < P (B)$
Answer dekho
Answer (D) $P (B ∣ A) < P (B)$
Dono sides ko $P (B)$ se multiply karne se dikhta hai ki overlap $P (A \cap B)$ , $P (A)$ guna $P (B)$ se kam hai. Ise $P (A)$ se divide karne par $P (B ∣ A) < P (B)$ milta hai: ek ke liye buri khabar doosre ke liye bhi buri khabar hai.
Quick answerMediumNCERT Miscellaneous Exercise, Q13
$A$ aur $B$ aise events hain jinke liye $P (A) \neq 0$ aur $P (A \cup B) = P (B)$ . Tab
$P (A ∣ B) = 1$ $P (B ∣ A) = 1$ $P (A ∣ B) = 0$ $P (B ∣ A) = 0$
Answer dekho
Answer (B) $P (B ∣ A) = 1$
Addition rule se, diya hua equation kehta hai ki $P (A)$ minus overlap zero hai. Isliye $P (A \cap B) = P (A)$ , aur $P (A)$ se divide karne par $P (B ∣ A) = 1$ milta hai.
14
Quick answerEasy
Bayes' theorem calculation ke steps ko order mein lagao.
1. priors, forward chances note karo
2. ek path ko P(A) se divide karo
3. paths jod kar P(A) nikaalo
4. hypotheses list karo
Answer dekho
Answer
hypotheses list karo
priors, forward chances note karo
paths jod kar P(A) nikaalo
ek path ko P(A) se divide karo
Priors aur forward probabilities ka koi matlab hone se pehle hypotheses ko partition banana chahiye; $P (A)$ denominator hai.
15
Quick answerEasy
Har rule ko uske formula se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
multiplication rule : $P (E) P (F ∣ E)$
independence : $P (E) P (F)$
F diya hone par complement : $1 - P (E ∣ F)$
total probability : $\sum_{j} P (E_{j}) P (A ∣ E_{j})$
Pehle do dono $P (E \cap F)$ ke barabar hain, doosra sirf independent events ke liye; teesra $P (E^{'} ∣ F)$ hai.
Quick answerEasy
Do mutually exclusive events, jinmein se har ek ki probability non-zero ho, hamesha hote hain.
Answer dekho
Answer dependent
Unke overlap ki probability 0 hai lekin unki probabilities ka product positive hai.
Quick answerMedium
Agar $P (A)$ aur $P (B)$ non-zero hain aur $P (A ∣ B) = P (A)$ , to $P (B ∣ A) = P (B)$ .
SahiGalat
Answer dekho
Answer Sahi
$P (A ∣ B) = P (A)$ ka matlab hai $P (A \cap B) = P (A) P (B)$ , jo $A$ aur $B$ mein symmetric hai, isliye $P (B ∣ A) = P (B)$ .

	Green	Red	White
K	2	5	3
L	4	4	2
M	7	1	2
N	0	3	7

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 13.1, Q1	Practice 13.1 · Q1
NCERT Exercise 13.1, Q2	Practice 13.1 · Q2
NCERT Exercise 13.1, Q3	Practice 13.1 · Q3
NCERT Exercise 13.1, Q4	Practice 13.1 · Q4
NCERT Exercise 13.1, Q5	Practice 13.1 · Q5
NCERT Exercise 13.1, Q6	Practice 13.1 · Q6
NCERT Exercise 13.1, Q7	Practice 13.1 · Q7
NCERT Exercise 13.1, Q8	Practice 13.1 · Q8
NCERT Exercise 13.1, Q9	Practice 13.1 · Q9
NCERT Exercise 13.1, Q10	Practice 13.1 · Q10
NCERT Exercise 13.1, Q11	Practice 13.1 · Q11
NCERT Exercise 13.1, Q12	Practice 13.1 · Q12
NCERT Exercise 13.1, Q13	Practice 13.1 · Q13
NCERT Exercise 13.1, Q14	Practice 13.1 · Q14
NCERT Exercise 13.1, Q15	Practice 13.1 · Q15
NCERT Exercise 13.1, Q16	Practice 13.1 · Q16
NCERT Exercise 13.1, Q17	Practice 13.1 · Q17
NCERT Exercise 13.2, Q1	Practice 13.2 · Q1
NCERT Exercise 13.2, Q2	Practice 13.2 · Q2
NCERT Exercise 13.2, Q3	Practice 13.2 · Q3
NCERT Exercise 13.2, Q4	Practice 13.2 · Q4
NCERT Exercise 13.2, Q5	Practice 13.2 · Q5
NCERT Exercise 13.2, Q6	Practice 13.2 · Q6
NCERT Exercise 13.2, Q7	Practice 13.2 · Q7
NCERT Exercise 13.2, Q8	Practice 13.2 · Q8
NCERT Exercise 13.2, Q9	Practice 13.2 · Q9
NCERT Exercise 13.2, Q10	Practice 13.2 · Q10
NCERT Exercise 13.2, Q11	Practice 13.2 · Q11
NCERT Exercise 13.2, Q12	Practice 13.2 · Q12
NCERT Exercise 13.2, Q13	Practice 13.2 · Q13
NCERT Exercise 13.2, Q14	Practice 13.2 · Q14
NCERT Exercise 13.2, Q15	Practice 13.2 · Q15
NCERT Exercise 13.2, Q16	Practice 13.2 · Q16
NCERT Exercise 13.2, Q17	Practice 13.2 · Q17
NCERT Exercise 13.2, Q18	Practice 13.2 · Q18
NCERT Exercise 13.3, Q1	Practice 13.3 · Q1
NCERT Exercise 13.3, Q2	Practice 13.3 · Q2
NCERT Exercise 13.3, Q3	Practice 13.3 · Q3
NCERT Exercise 13.3, Q4	Practice 13.3 · Q4
NCERT Exercise 13.3, Q5	Practice 13.3 · Q5
NCERT Exercise 13.3, Q6	Practice 13.3 · Q6
NCERT Exercise 13.3, Q7	Practice 13.3 · Q7
NCERT Exercise 13.3, Q8	Practice 13.3 · Q8
NCERT Exercise 13.3, Q9	Practice 13.3 · Q9
NCERT Exercise 13.3, Q10	Practice 13.3 · Q10
NCERT Exercise 13.3, Q11	Practice 13.3 · Q11
NCERT Exercise 13.3, Q12	Practice 13.3 · Q12
NCERT Exercise 13.3, Q13	Practice 13.3 · Q13
NCERT Exercise 13.3, Q14	Practice 13.3 · Q14
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercise · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercise · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercise · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercise · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercise · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercise · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous exercise · Q11
NCERT Miscellaneous Exercise, Q12	Miscellaneous exercise · Q12
NCERT Miscellaneous Exercise, Q13	Miscellaneous exercise · Q13

Quick check: MCQs

13.1 Introduction2 questions
13.2.1 Conditional probability ki properties4 questions
13.3 Probability par Multiplication Theorem3 questions
13.4 Independent Events4 questions
Bayes' theorem4 questions
Random variables2 questions
Miscellaneous Examples2 questions

Glossary · 12

Key terms (English + Hindi)

Bayes' theoremबेज़-प्रमेयbez prameya: Woh rule ki posterior $P (E_{i} ∣ A)$ barabar hai $P (E_{i}) P (A ∣ E_{i})$ ko $P (A)$ se divide karne ke: effect dikhne ke baad cause ka chance.Chapter mein pehli baar yahan aaya →

conditional probabilityसप्रतिबंध प्रायिकताsapratibandh praayikta: $F$ ke hone ka pata chalne ke baad $E$ ki probability: $P (E ∣ F) = P (E \cap F) / P (F)$ , jahan $P (F) \neq 0$ .Chapter mein pehli baar yahan aaya →

hypothesisपरिकल्पनाparikalpana: Bayes' theorem mein partition ka ek event $E_{i}$ : jo dikha uska ek possible cause.Chapter mein pehli baar yahan aaya →

independent eventsस्वतंत्र घटनाएँsvatantra ghatnaayen: Aise events jinke liye $P (E \cap F) = P (E) P (F)$ , yaani ek ko jaanne se doosre ka chance nahi badalta. Warna woh dependent hain.

multiplication ruleप्रायिकता का गुणन नियमpraayikta ka gunan niyam: $P (E \cap F) = P (E) P (F ∣ E)$ : dono ka chance pehle ka chance guna pehle diye hone par doosre ka chance hai.Chapter mein pehli baar yahan aaya →
mutually exclusive eventsपरस्पर अपवर्जी घटनाएँparaspar apavarji ghatnaayen: Aise events jinka koi outcome common nahi, $E \cap F = \emptyset$ ; yeh saath mein nahi ho sakte.Chapter mein pehli baar yahan aaya →

partition of a sample spaceप्रतिदर्श समष्टि का विभाजनpratidarsh samashti ka vibhaajan: Events $E_{1}, \dots, E_{n}$ jo pairwise disjoint hon, milkar $S$ banayein, aur har ek ki probability non-zero ho.
posterior probability: $P (E_{i} ∣ A)$ , evidence $A$ ko dhyaan mein lene ke baad kisi hypothesis ki probability.Chapter mein pehli baar yahan aaya →
prior probability: $P (E_{i})$ , evidence ko dhyaan mein lene se pehle kisi hypothesis ki probability.Chapter mein pehli baar yahan aaya →

random variableयादृच्छिक चरyaadrichchhik char: Ek rule jo random experiment ke har outcome ke liye ek real number deta hai: $S$ se real numbers tak ek function.Chapter mein pehli baar yahan aaya →

theorem of total probabilityसंपूर्ण प्रायिकता प्रमेयsampoorn praayikta prameya: Partition $E_{1}, \dots, E_{n}$ ke liye: $P (A) = \sum_{j} P (E_{j}) P (A ∣ E_{j})$ .
tree diagramवृक्ष आरेखvriksh aarekh: Stages wale experiment ki branches wali picture; path ke saath multiply karne se us path ki probability milti hai.Chapter mein pehli baar yahan aaya →

Probability: Class 12 Maths Chapter 13 Hinglish mein

Is chapter mein aap kya seekhenge

13.1 Introduction

13.2 Conditional Probability

13.2.1 Conditional probability ki properties

Practice 13.1: answers ke saath

13.3 Probability par Multiplication Theorem

13.4 Independent Events

Practice 13.2: answers ke saath

13.5 Bayes' Theorem

13.5.1 Sample space ka partition

13.5.2 Total probability ka theorem

Bayes' theorem

Random variables

Practice 13.3: answers ke saath

Miscellaneous Examples

Summary aur key points

Chapter 13 par Miscellaneous Exercise

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye