Conditional probability — P(A - B) = P(A∩B) - P(A): Question bank

2.7.8 · D5 · Maths › Statistics & Probability — Intermediate › Conditional probability — P(A - B) = P(A∩B) - P(A)

Notation reminder, taaki kuch bhi use hone se pehle unnamed na rahe:

$A \cap B$ ::: " $A$ aur $B$ " dono hote hain — dono events ka overlap.
$A^{c}$ ::: "not $A$ " — $A$ ke bahar ke saare outcomes (uska complement).
$B^{c}$ ::: "not $B$ ".
$P (A ∣ B)$ ::: $A$ ki probability given $B$ , define hoti hai sirf jab $P (B) > 0$ ; bar ko "given" ki tarah padho.
$n (\cdot)$ ::: kisi event mein outcomes ki count (sirf neeche equally-likely picture mein use hogi).

True or false — justify

P (A ∣ B) = P (B ∣ A)

always.

False. Dono ka numerator

P (A \cap B)

share hota hai, lekin pehla

P (B)

se divide hota hai aur doosra

P (A)

se (har ek ki apni positivity requirement ke saath); jab tak

P (A) = P (B)

na ho yeh alag hote hain.

A

and

B

are independent then

P (A ∣ B) = P (A)

(assuming

P (B) > 0

True. Independence ka matlab hai ki

B

jaanno toh

A

ke baare mein kuch naya nahi milta, toh conditioned probability unconditioned ke barabar hoti hai — yahi Independence of events ke through definition hai.

P (A ∣ B) + P (A^{c} ∣ B) = 1

always (when

P (B) > 0

True.

B

ko universe maan lene par,

A

aur

A^{c}

us universe ko completely split karte hain, toh unki conditional probabilities ka sum 1 hona hi chahiye.

P (A ∣ B) + P (A ∣ B^{c}) = 1

always.

False. Bar ke baad wala event change nahi kar sakte;

B

aur

B^{c}

do alag universes hain (har ek ki apni positive probability chahiye), toh in dono numbers ka 1 mein add hone ki koi wajah nahi.

B \subseteq A

(every

B

-outcome is also in

A

) and

P (B) > 0

then

P (A ∣ B) = 1

True. Tab

A \cap B = B

, toh

P (A ∣ B) = P (B) / P (B) = 1

— ek baar

B

ke andar gaye, toh

A

guaranteed hai.

A \subseteq B

and

P (B) > 0

then

P (A ∣ B) = 1

False. Yahan

A \cap B = A

, toh

P (A ∣ B) = P (A) / P (B)

, jo generally 1 se kam hoti hai; bade set

B

ke andar hona chhote

A

ko force nahi karta.

P (A ∣ B)

can be larger than

P (A)

(when

P (B) > 0

True. Conditioning probability ko badha sakti hai: agar

B

A

zyada likely ho jaata hai (positive association), toh

P (A ∣ B) > P (A)

P (A ∣ B)

is always at least as large as

P (A \cap B)

(when

P (B) > 0

True.

P (B) \leq 1

se divide karne par woh same rehti hai ya barhti hai, kyunki

P (A ∣ B) = P (A \cap B) / P (B) \geq P (A \cap B)

A

and

B

are mutually exclusive (never both happen) and

P (B) > 0

, then

P (A ∣ B) = 0

True. Mutually exclusive ka matlab

A \cap B = \emptyset

, toh

P (A \cap B) = 0

aur ratio

0

hai —

B

ke andar,

A

kabhi nahi ho sakta.

Mutually exclusive events with positive probability are independent.

False. Agar woh ek doosre ko exclude karte hain, toh

P (A ∣ B) = 0 \neq = P (A)

, toh

B

jaanno toh

A

ka chance badal jaata hai — yeh dependence hai, independence nahi.

Spot the error

P (A ∣ B) = \frac{P ( A \cap B )}{P ( A )}

Galat denominator: given event

P (B)

se divide karna chahiye (jo positive hona chahiye),

P (A)

se nahi.

P (A)

se divide karne par silently

P (B ∣ A)

compute ho jaata hai.

"Ek test

99%

accurate hai, maine positive test kiya, toh main

99%

likely sick hoon."

Woh

99%

P (+ ∣ sick)

hai, jo tumhein chaahiye uska ulta,

P (sick ∣ +)

. Bayes' Theorem ke bina bar ko flip karna yeh ignore karta hai ki disease kitni rare hai (false positives dominate kar sakte hain).

P (A \cap B) = P (A) \cdot P (B)

, toh sab events aise hi multiply hote hain."

Sirf tab jab

A, B

independent hon. General Multiplication rule of probability hai

P (A \cap B) = P (B) P (A ∣ B)

(

P (B) > 0

ke liye); plain product ek special case hai.

"Kyunki

P (A ∣ B) = 0.7

aur

P (A ∣ B^{c}) = 0.7

hai, toh

P (A) = 1.4

Conditionals ko aise add nahi kar sakte. Sahi combination hai Law of total probability:

P (A) = P (A ∣ B) P (B) + P (A ∣ B^{c}) P (B^{c})

, ek weighted average jo

\leq 1

rehta hai.

P (B) = 0

, toh

P (A ∣ B) = \frac{P ( A \cap B )}{0} = \infty

Infinity nahi — yeh undefined hai. Ek impossible event par condition nahi kar sakte; formula ko

P (B) > 0

chahiye aur warna yeh simply apply nahi hota.

"Bina replacement ke do kings:

P = \frac{4}{52} \cdot \frac{4}{52}

Doosri draw ko pehli par condition karna padega: ek king remove karne ke baad,

51

cards mein sirf

3

kings bachte hain, toh yeh

\frac{4}{52} \cdot \frac{3}{51}

hai.

"Ek tree mein, do alag paths combine karne ke liye main ek branch ke along bhi multiply karta hoon AUR upar bhi."

Branch ke along multiply karo (conditionals chain karo), lekin alag branches ko same outcome mein combine karte waqt add karo.

Edge cases

Agar

P (B) = 1

(B certain hai), toh

P (A ∣ B)

kya hai?

Yeh

P (A)

ke barabar hai, kyunki "universe"

B

pehle se hi sab kuch hai — ek sure event par condition karne se kuch nahi badalta.

Agar

A = B

(same event,

P (A) > 0

) ho, toh

P (A ∣ B)

kya hai?

P (A ∣ A) = P (A \cap A) / P (A) = P (A) / P (A) = 1

— kisi event ko given maano, toh woh event definitely hold karti hai.

Agar

P (A) = 0

lekin

P (B) > 0

ho, toh

P (A ∣ B)

kya hai?

P (A \cap B) \leq P (A) = 0

, toh numerator

0

hai aur

P (A ∣ B) = 0

; ek impossible

A

kisi bhi universe ke andar impossible rehta hai.

Agar

A

aur

B

independent hain aur

P (A) = 1

hai, toh kya

P (A ∣ B) = 1

hai?

Haan:

P (A ∣ B) = P (A) = 1

; ek certain event certain rehta hai, aur independence trivially satisfy hoti hai.

Kya

P (A ∣ B) = P (A)

hold ho sakta hai jab

A

aur

B

mutually exclusive hon?

Sirf tab agar

P (A) = 0

ho, kyunki exclusivity

P (A ∣ B) = 0

force karti hai, jo

P (A)

ke barabar sirf tab ho sakta hai jab

A

khud impossible ho.

Replacement ke saath

P (2nd king ∣ 1st king)

ka kya hoga without ki jagah?

Universe reset ho jaata hai ek full

52

-card deck pe, toh yeh

4/52

ho jaata hai — draws tab independent hote hain aur condition matter karna band kar deti hai.

Agar ek tree branch mein

P (B)

bahut chota hai lekin

P (A ∣ B)

bahut bada hai, toh kya path

P (A \cap B)

bada hai?

Zaroor nahi — path probability product

P (B) P (A ∣ B)

hai, toh chota

P (B)

ek near-certain conditional ko bhi shrink kar sakta hai.

Recall Ek-line survival guide

Bar ke daayein wale event — given event $B$ — se divide karo — aur sirf tab jab $P (B) > 0$ ho; bar ke baayein wala swap kar sakte ho (complement law $P (A ∣ B) + P (A^{c} ∣ B) = 1$ ) lekin kabhi daayein wala nahi.

Connections

Independence of events — $P (A ∣ B) = P (A)$ boundary jise kai traps probe karte hain.
Multiplication rule of probability — isliye doosra card $3/51$ hai, $4/52$ nahi.
Bayes' Theorem — $P (+ ∣ sick)$ ko $P (sick ∣ +)$ mein flip karne ka sahi tarika.
Law of total probability — isliye $P (A ∣ B) + P (A ∣ B^{c}) \neq = 1$ lekin ek weighted sum $P (A)$ recover karta hai.
Sample space and events — woh universe jo conditioning karte waqt shrink hota hai.
Tree diagrams — branches ke along multiply karo, unke across add karo.

Question bank — Conditional probability — P(A - B) = P(A∩B) - P(A)

True or false — justify

Spot the error

Why questions

Edge cases

Connections