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

2.7.8 · D4 · Maths › Statistics & Probability — Intermediate › Conditional probability — P(A - B) = P(A∩B) - P(A)

Shuru karne se pehle, ek picture jo poori idea ko dimaag mein fix kar de: conditioning universe ko shrink karti hai.

Figure — Conditional probability — P(A - B) = P(A∩B) - P(A)

Bada peach rectangle poora sample space hai. Jab aapko bata diya jaaye ki $B$ hua, toh aap violet blob $B$ ke andar band ho jaate ho. $A$ ka sirf wahi hissa count karta hai jo magenta overlap $A \cap B$ hai. $P (A ∣ B)$ bas yahi measure karta hai.

Level 1 — Recognition

Yahan aapko bas pieces ka naam lena hai aur formula ko sahi se padhna hai.

Recall Solution L1.1

KYA: Formula hai $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . KYUN: $B$ given event hai, isliye yeh denominator hai (naya universe). $P (A ∣ B) = \frac{0.12}{0.30} = 0.4$ Check: $0.4 < 1$ , jaisa kisi bhi probability ko hona chahiye. ✔

Recall Solution L1.2

Jawab: undefined. Aap kabhi bhi $P (B) = 0$ se divide nahi kar sakte. Ek impossible event par conditioning ka koi matlab nahi — andar baithne ke liye koi shrunken universe hi nahi hai. Formula require karta hai $P (B) > 0$ .

Recall Solution L1.3

Bar ke daayein wala event $> 3$ hai, isliye hum $P (> 3)$ se divide karte hain. ( $> 3 = {4, 5, 6}$ , toh $P (> 3) = \frac{3}{6} = \frac{1}{2}$ .) Bar ka dost neeche jaata hai.

Level 2 — Application

Ab real counts ko formula mein plug karo.

Recall Solution L2.1

Step 1 — overlap. $A \cap B = {3, 6}$ , toh $P (A \cap B) = \frac{2}{6}$ . Step 2 — condition. $P (B) = \frac{4}{6}$ . Step 3 — divide. $P (A ∣ B) = \frac{2/6}{4/6} = \frac{2}{4} = \frac{1}{2}$ Naye universe ke andar sanity check: ${3, 4, 5, 6}$ mein se, $3$ ke multiples hain ${3, 6}$ — yeh $4$ mein se $2$ hai. ✔

Recall Solution L2.2

$26$ red cards hain. multiplication rule use karo $P (R_{1} \cap R_{2}) = P (R_{1}) P (R_{2} ∣ R_{1})$ . Step 1: $P (R_{1}) = \frac{26}{52} = \frac{1}{2}$ . Step 2: Ek red jaane ke baad, $51$ cards mein $25$ reds bachte hain, toh $P (R_{2} ∣ R_{1}) = \frac{25}{51}$ . Step 3: $P (both red) = \frac{26}{52} \cdot \frac{25}{51} = \frac{25}{102}$

Recall Solution L2.3

KYA hum condition kar rahe hain: pehli draw magenta thi, toh ek magenta gayi. Naya universe: $7$ marbles bache hain — $2$ magenta, $5$ orange. $P (2nd orange ∣ 1st magenta) = \frac{5}{7}$ Bas yahi matter karta hai ki kya bacha hai, na ki jo colour hataaya woh apne future mein kya karega.

Level 3 — Analysis

Yahan aapko sahi tool choose karna hai: complement, multiplication, ya independence check.

Recall Solution L3.1

Sahi complement law condition ko fixed rakhta hai: $P (A ∣ B) + P (A^{c} ∣ B) = 1 ⟹ P (A^{c} ∣ B) = 1 - 0.7 = 0.3$ Yeh kyun, na ki $P (A ∣ B^{c})$ : fixed universe $B$ ke andar, events $A$ aur $A^{c}$ use poora split karte hain, isliye unki conditional probabilities $1$ tak add hoti hain. Condition ko $B^{c}$ mein badalna poori tarah alag universe mein move karna hai.

Recall Solution L3.2

Tool choice: independence test use karo $P (A \cap B) = ? P (A) P (B)$ (dekho Independence of events). $P (A) P (B) = 0.4 \times 0.5 = 0.20 = P (A \cap B)$ Yeh independent hain. Equivalently $P (A ∣ B) = \frac{0.2}{0.5} = 0.4 = P (A)$ — $B$ jaanna kuch bhi nahi badla.

Recall Solution L3.3

Tool choice: "bike-owners mein se, $25%$ ..." ek conditional hai $P (S ∣ K) = 0.25$ . Multiplication use karo: $P (K \cap S) = P (K) P (S ∣ K) = 0.60 \times 0.25 = 0.15$ Toh poori class ka $15%$ dono ke malik hain.

Level 4 — Synthesis

Multiplication, total probability, aur condition ko reverse karna combine karo.

Recall Solution L4.1

Tool: Law of total probability $D$ ko partition ${M_{1}, M_{2}}$ par split karta hai: $P (D) = P (M_{1}) P (D ∣ M_{1}) + P (M_{2}) P (D ∣ M_{2})$ $= 0.70 \times 0.02 + 0.30 \times 0.05 = 0.014 + 0.015 = 0.029$ Toh saare bolts ka $2.9%$ defective hai. Upar wala tree padho: har path neeche multiply karta hai, phir hum do "defective" leaves ko add karte hain.

Recall Solution L4.2

Tool: Bayes' Theorem condition ko sahi se flip karta hai: $P (M_{2} ∣ D) = \frac{P ( M _{2} ) P ( D ∣ M _{2} )}{P ( D )} = \frac{0.30 \times 0.05}{0.029} = \frac{0.015}{0.029} = \frac{15}{29} \approx 0.517$ Bhale hi $M_{2}$ sirf $30%$ bolts banati hai, phir bhi woh aadhe se zyada defects ke liye zimmedaar hai — kyunki uski defect rate zyada hai.

Recall Solution L4.3

Step 1 — $+$ ki total probability: $P (+) = 0.02 \times 0.95 + 0.98 \times 0.10 = 0.019 + 0.098 = 0.117$ Step 2 — Bayes: $P (sick ∣ +) = \frac{0.02 \times 0.95}{0.117} = \frac{0.019}{0.117} \approx 0.1624$ Itna kam kyun? Bimari rare hai, isliye zyaatar positives huge healthy group ke false positives hain. Sahi tarike se conditioning aapko un sabhi tak restrict karti hai jinhone $+$ test kiya, phir beemar fraction nikalti hai — lagbhag $16%$ , $95%$ nahi.

Level 5 — Mastery

Scratch se poora scenario model karo.

Recall Solution L5.1

Har urn $\frac{1}{3}$ probability se choose hoti hai, aur har urn mein $5$ balls hain. $P (M ∣ U_{1}) = \frac{2}{5}, P (M ∣ U_{2}) = \frac{4}{5}, P (M ∣ U_{3}) = \frac{1}{5}$ (a) Law of total probability: $P (M) = \frac{1}{3} (\frac{2}{5} + \frac{4}{5} + \frac{1}{5}) = \frac{1}{3} \cdot \frac{7}{5} = \frac{7}{15}$ (b) Bayes: $P (U_{2} ∣ M) = \frac{\frac{1}{3} \cdot \frac{4}{5}}{\frac{7}{15}} = \frac{4/15}{7/15} = \frac{4}{7}$ Urn $2$ magenta mein sabse ameer hai, isliye magenta dekhne se woh sabse likely source ban jaata hai — aadhe se zyada.

Recall Solution L5.2

Strategy: "at least one" apne complement "no ace" se aasaan hota hai. Conditional no-ace draws chain karo (har draw deck ko shrink karti hai). $P (no ace) = \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50}$ $= \frac{48 \cdot 47 \cdot 46}{52 \cdot 51 \cdot 50} = \frac{103776}{132600} = \frac{4324}{5525}$ $P (at least one ace) = 1 - \frac{4324}{5525} = \frac{1201}{5525} \approx 0.2174$

Recall Solution L5.3

Sample space: ${H H, H T, T H, T T}$ , har ek probability $\frac{1}{4}$ ke saath (dekho Sample space and events). $A = {H H, H T}$ toh $P (A) = \frac{1}{2}$ . $B = {H T, T H}$ toh $P (B) = \frac{1}{2}$ . $A \cap B = {H T}$ toh $P (A \cap B) = \frac{1}{4}$ . Independence test: $P (A) P (B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = P (A \cap B)$ — independent! Conditional: $P (A ∣ B) = \frac{1/4}{1/2} = \frac{1}{2} = P (A)$ . "Exactly ek head" jaanna koi clue nahi deta ki woh flip kaunsa tha.

Recall Feynman recap: poori ladder ek saans mein

Yahan har problem same move hai: koi aapko kuch bataata hai, toh aapki possibilities ka box shrink ho jaata hai. L1 bas pieces ka naam leta hai. L2 chote box ke andar count karta hai. L3 choose karta hai ki complement karo, multiply karo, ya independence test karo. L4 ek tree neeche chalta hai, paths add karta hai (total probability), aur arrows reverse karta hai (Bayes). L5 yeh sab ek saath karta hai. Denominator hamesha wahi hota hai jo aapko bataya gaya tha.

Connections

Multiplication rule of probability — har "without replacement" chain ko power deta hai (L2, L5).
Law of total probability — ek partition par conditionals ko sum karta hai (L4, L5).
Bayes' Theorem — condition ko reverse karta hai (L4.2, L4.3, L5.1b).
Independence of events — arithmetic test (L3.2, L5.3).
Sample space and events — woh universe jo hum shrink karte hain.
Tree diagrams — L4/L5 steps ka visual chaining.