Worked examples — Mutually exclusive events — addition rule

2.7.6 · D3 · Maths › Statistics & Probability — Intermediate › Mutually exclusive events — addition rule

Yahan use hone wala har symbol parent note mein define kiya gaya hai. Do main rules ka quick refresher:

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Scenario matrix

Har addition-rule problem theek inhi case classes mein se ek mein aati hai. Table mein sab list hain, aur neeche ke examples us cell ke saath label kiye gaye hain jisme woh fit hote hain.

Cell	Case class	Isme tricky kya hai	Example
C1	Disjoint events, bas add karo	kuch nahi — friendly case	Ex 1
C2	Overlapping events, subtract karna padega	double-count trap	Ex 2
C3	Teen ya zyada pairwise-disjoint events	sum $\sum P(A_i)$	Ex 3
C4	Complement route ($P(A^{\prime })=1-P(A)$)	"not" wali phrasing	Ex 4
C5	Degenerate: ek impossible ($P=0$) ya certain ($P=1$) event	zero / one edge inputs	Ex 5
C6	Impossibility check: $P(A)+P(B)>1$	kya woh disjoint ho bhi sakte hain?	Ex 6
C7	Real-world word problem	words ko sets mein translate karna	Ex 7
C8	Exam twist: unknown overlap ke liye ulta solve karna	general rule ko rearrange karna	Ex 8

Hum poore time Venn diagrams use karte hain — ek rectangle $S$ ke andar do overlapping ovals. Dekho Venn diagrams in probability.

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Example 1 — Disjoint, bas add karo (Cell C1)

Step 1 — Events ko sets ki tarah likho. $A=\{1\}$, $B=\{6\}$. Yeh step kyun? Words ko sample space $S=\{1,2,3,4,5,6\}$ ke subsets mein badalne se hum dekh sakte hain ki woh overlap karte hain ya nahi. Dekho Probability — basic definitions & sample space.

Step 2 — Overlap check karo. $A\cap B=\varnothing$ — koi bhi face ek saath 1 aur 6 nahi ho sakta. Yeh step kyun? Yahi woh sawaal hai jo parent note humse poochne ki request karta hai: kya dono ek saath ho sakte hain? Nahi → subtraction term zero hai.

Step 3 — Add karo. Har face equally likely hai, isliye $P(A)=\frac{1}{6}$, $P(B)=\frac{1}{6}$, aur $P(A\cup B)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}.$ Yeh step kyun? Zero overlap ke saath special rule apply hota hai — seedha addition.

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Example 2 — Overlapping, subtract karna zaroori (Cell C2)

Step 1 — Events ka naam rakho. $A=\{\text{hearts}\}$, $B=\{\text{face cards}\}$. Yeh step kyun? Humhe sizes chahiye aur, sabse zaroori, overlap chahiye.

Step 2 — Overlap nikalo. Cards jo dono heart bhi hain aur face card bhi: Jack, Queen, King of hearts — yeh $3$ cards hain. Toh $P(A\cap B)=\frac{3}{52}$. Yeh step kyun? Yeh teen cards Venn diagram ke dono ovals mein rehte hain (Fig 1). $P(A)+P(B)$ add karne se inhe do baar count kiya jaata hai.

Step 3 — General rule apply karo. $P(A\cup B)=\frac{13}{52}+\frac{12}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}.$ Yeh step kyun? Hum pieces add karte hain, phir double-counted overlap ko ek baar remove karte hain — inclusion–exclusion. Dekho General addition rule & inclusion–exclusion.

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Example 3 — Teen pairwise-disjoint events (Cell C3)

Step 1 — Pairwise disjoint confirm karo. Ek single spin exactly ek colour deta hai, isliye do colour-events ek saath nahi ho sakte: har pairwise intersection $\varnothing$ hai. Yeh step kyun? $n$-event rule $P(A_1\cup \cdots \cup A_n)=\sum P(A_i)$ sirf tab apply hota hai jab events pairwise mutually exclusive hon.

Step 2 — Teen probabilities ka sum karo. $P=0.4+0.25+0.2=\mathrm{0.85.}$ Yeh step kyun? Jab har overlap zero ho, toh har subtraction term vanish ho jaati hai — union ek seedhe sum mein collapse ho jaata hai.

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Example 4 — Complement route (Cell C4)

Step 1 — Event aur uska complement identify karo. Maano $A=\{\text{picks correctly}\}$, toh $A^{\prime }=\{\text{does not pick correctly}\}$. Yeh step kyun? "Not" wali phrasing seedha complement ki taraf point karti hai. $A$ aur $A^{\prime }$ mutually exclusive hain aur milke $S$ fill karte hain (exhaustive). Dekho Complementary events.

Step 2 — Complement corollary use karo. Kyunki $P(A)+P(A^{\prime })=1$, $P(A^{\prime })=1-P(A)=1-\frac{1}{5}=\frac{4}{5}.$ Yeh step kyun? Chaar galat options seedhe count karna bhi kaam karta hai, lekin complement ek clean subtraction hai — kaafi kam error-prone.

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Example 5 — Degenerate inputs: zero aur certain (Cell C5)

Step 1 — Edge probabilities assign karo. Koi face 7 nahi dikhata, isliye $A=\varnothing$ aur $P(A)=0$. Har face $\le 6$ hai, isliye $B=S$ aur $P(B)=1$. Yeh step kyun? Yeh probability ke boundary values hain: $0\le P\le 1$. Yeh test karte hain ki rule phir bhi sahi behave karta hai ya nahi.

Step 2 — Union. Kyunki $A=\varnothing$, "$A$ ya $B$" simply $B$ hai: $P(A\cup B)=P(\varnothing \cup S)=P(S)=1.$ Rule se check karo: $A\cap B=\varnothing \cap S=\varnothing$, toh $P(A\cup B)=0+1-0=1.$ ✓ Yeh step kyun? Ek impossible event add karne se kuch nahi badalta — uski probability $0$ contribute karti hai.

Step 3 — Intersection. $A\cap B=\varnothing \cap S=\varnothing$, toh $P(A\cap B)=0$. Yeh step kyun? Ek impossible event kisi bhi cheez ke saath koi outcome share nahi karta, isliye $A$ aur $B$ yahan (trivially) mutually exclusive hain.

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Example 6 — Kya woh disjoint ho bhi sakte hain? (Cell C6)

Step 1 — Disjoint assumption test karo. Agar mutually exclusive hain, toh $P(\text{rain}\cup \text{wind})=0.7+0.5=1.2$. Yeh step kyun? Koi bhi probability, including union, $P\le 1$ satisfy karni chahiye. $1.2$ ki value forbidden hai.

Step 2 — Conclude karo ki unhe overlap karna hi hai. Kyunki $1.2>1$, woh disjoint nahi ho sakte. General rule use karo constraint $P(\text{rain}\cup \text{wind})\le 1$ ke saath: $P(\text{rain}\cap \text{wind})=P(\text{rain})+P(\text{wind})-P(\text{rain}\cup \text{wind})\ge 0.7+0.5-1=\mathrm{0.2.}$ Yeh step kyun? General rule ko rearrange karne se "$\le 1$" ceiling overlap par ek floor ban jaata hai.

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Example 7 — Real-world word problem (Cell C7)

Step 1 — Probabilities mein translate karo. $n(S)=30$ ke saath: $P(F)=\frac{18}{30},P(Sp)=\frac{12}{30},P(F\cap Sp)=\frac{5}{30}.$ Yeh step kyun? Equally-likely students ka matlab hai probability = (count)/(30).

Step 2 — General rule apply karo (woh overlap karte hain). $P(F\cup Sp)=\frac{18}{30}+\frac{12}{30}-\frac{5}{30}=\frac{25}{30}=\frac{5}{6}.$ Yeh step kyun? 5 dono-language students Venn diagram ke overlap mein baithte hain; subtract kiye bina add karne se unhe double-count kiya jaata.

Step 3 — "Neither" complement se. $P(\text{neither})=1-P(F\cup Sp)=1-\frac{25}{30}=\frac{5}{30}=\frac{1}{6}.$ Yeh step kyun? "Kam se kam ek padhta hai" aur "koi nahi padhta" complementary hain.

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Example 8 — Exam twist: ulta solve karo (Cell C8)

Step 1 — General rule ko overlap ke liye rearrange karo. $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.55+0.4-0.8=\mathrm{0.15.}$ Yeh step kyun? General rule mein chaar quantities hain; koi bhi teen given hon, algebra chautha de deta hai. Yeh exam ka favourite move hai.

Step 2 — Mutually exclusive? Nahi: $P(A\cap B)=0.15\ne 0$, toh overlap hai. Yeh step kyun? Mutual exclusivity exactly yeh statement hai ki $P(A\cap B)=0$; nonzero overlap ise rule out karta hai.

Step 3 — Independent? Independence ke liye humein chahiye hoga $P(A\cap B)=P(A)P(B)$. Check karo: $0.55\times 0.4=0.22\ne 0.15$. Toh woh independent bhi nahi hain. Dekho Independent events — multiplication rule aur Conditional probability. Yeh step kyun? Independence intersection par ek multiplication test hai — exclusivity se (jo ek addition-side condition hai) bilkul alag condition.

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Coverage check — kya humne har cell hit ki?

Recall Matrix cell → example

C1 disjoint add ::: Example 1 (die, 1 or 6) C2 overlapping subtract ::: Example 2 (heart or face card) C3 teen pairwise-disjoint ::: Example 3 (spinner colours) C4 complement route ::: Example 4 (wrong guess) C5 degenerate 0 / 1 inputs ::: Example 5 (roll a 7 / roll $\le 6$) C6 sum > 1 impossibility ::: Example 6 (rain & wind) C7 real-world word problem ::: Example 7 (French / Spanish) C8 exam twist, ulta solve karo ::: Example 8 (overlap nikalo)

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Connections

• Mutually exclusive events — addition rule (parent)
• General addition rule & inclusion–exclusion
• Complementary events
• Independent events — multiplication rule
• Conditional probability
• Venn diagrams in probability
• Probability — basic definitions & sample space