Exercises — Mutually exclusive events — addition rule

Kya dono ek saath ho sakte hain? Ek roll mein ek hi face aata hai; woh ek saath $3$ aur $6$ nahi ho sakta. Toh $A\cap B=\varnothing$ — yeh mutually exclusive hain. $P(A)=\frac{1}{6}$, $P(B)=\frac{1}{6}$. Koi overlap nahi, toh: $P(A\cup B)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}.$

Kya dono ek saath ho sakte hain? Haan — King of Hearts ek heart bhi hai aur King bhi. Toh $A\cap B=\{\text{King of Hearts}\}\ne \varnothing$. Isliye $A$ aur $B$ mutually exclusive nahi hain. (Inhe combine karne ke liye general rule lagana padega, $P(A\cap B)=\frac{1}{52}$ subtract karke.)

Yeh teen alag faces hain, jinmein se koi bhi do ek saath nahi aa sakte — pairwise mutually exclusive. Pairwise exclusive events ke liye bas add karo: $P=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}.$

Kya dono ho sakte hain? Haan — Ace of Spades. Exclusive nahi, toh general rule use karo. $P(\text{Ace})=\frac{4}{52}$, $P(\text{Spade})=\frac{13}{52}$, overlap $P(\text{Ace}\cap \text{Spade})=\frac{1}{52}$. $P=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}.$ Count karke check karo: 4 Aces + 13 Spades − 1 double-counted (Ace of Spades) = $16$ favourable cards. ✓

Colours exclusive hain (ek ball exactly ek colour ki hoti hai), toh add karo: $P(\text{red}\cup \text{green})=0.4+0.25=\mathrm{0.65.}$ "Not blue" ke liye complement rule use karo $P(A^{\prime })=1-P(A)$: $P(\text{not blue})=1-0.35=\mathrm{0.65.}$ (Same number — kyunki "not blue" yahan "red or green" hi hai, kyunki teen colours exhaustive hain.)

Overlap nikalne ke liye general rule ko rearrange karo: $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.45+0.35-0.8=0.$ Overlap $0$ hai, toh $A\cap B=\varnothing$ — yeh mutually exclusive hain. (Jab $P(A)$ aur $P(B)$ ko seedha add karne par exactly $P(A\cup B)$ aaye, toh subtract karne ke liye kuch tha hi nahi.)

$P(A\cap B)=0.5+0.4-0.7=\mathrm{0.2.}$ Overlap $0.2\ne 0$ hai, toh yeh mutually exclusive nahi hain — yeh $20\%$ probability share karte hain.

Agar yeh exclusive hote, toh $P(A\cup B)=0.55+0.6=1.15>1$ — yeh impossible hai, kyunki probability $1$ se zyada nahi ho sakti. Toh yeh mutually exclusive nahi ho sakte. Union zyada se zyada $1$ ho sakta hai, toh $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ se overlap tab sabse chhota hoga jab $P(A\cup B)$ sabse bada ($=1$) ho: $P(A\cap B{)}_{\min }=0.55+0.6-1=\mathrm{0.15.}$

Maano $F$ = football, $C$ = cricket. Diya gaya $P(F)=0.4$, $P(C)=0.3$, $P(F\cap C)=0.1$. (a) At least one = union. General rule (yeh overlap karte hain!): $P(F\cup C)=0.4+0.3-0.1=\mathrm{0.6.}$ (b) "Neither" "at least one" ka complement hai: $P(\text{neither})=1-P(F\cup C)=1-0.6=\mathrm{0.4.}$ (c) "Football only" = football minus shared part: $P(F\text{ only})=P(F)-P(F\cap C)=0.4-0.1=\mathrm{0.3.}$

Independent ka matlab hai "$A$ tumhe $B$ ke baare mein kuch nahi batata", jo multiplication rule se intersection fix karta hai: $P(A\cap B)=P(A)P(B)=0.5\times 0.4=\mathrm{0.2.}$ Phir general addition rule deta hai: $P(A\cup B)=0.5+0.4-0.2=\mathrm{0.7.}$ Mutually exclusive hone par $P(A\cap B)=0$ force hota, toh: $P(A\cup B)=0.5+0.4=\mathrm{0.9.}$ Alag answers — kyunki independence aur exclusivity overlap par alag constraints hain (ek ise $P(A)P(B)$ banata hai, doosra ise $0$ banata hai).

Exhaustive & exclusive ka matlab hai charon ka sum $1$ hoga: $0.3+0.25+P(\text{green})+P(\text{yellow})=1\Rightarrow P(\text{green})+P(\text{yellow})=\mathrm{0.45.}$ $P(\text{green})=2P(\text{yellow})$ substitute karo: $2P(\text{yellow})+P(\text{yellow})=0.45\Rightarrow 3P(\text{yellow})=0.45\Rightarrow P(\text{yellow})=\mathrm{0.15.}$ Toh $P(\text{green})=0.30$. Check: $0.3+0.25+0.30+0.15=1$. ✓ Red aur yellow exclusive hain, toh add karo: $P(\text{red}\cup \text{yellow})=0.3+0.15=\mathrm{0.45.}$

Outcomes count karo ($n(S)=30$):

• Multiples of $3$: $3,6,\dots ,30$ → $10$ numbers, $P(A)=\frac{10}{30}$.
• Multiples of $5$: $5,10,\dots ,30$ → $6$ numbers, $P(B)=\frac{6}{30}$.
• Dono = multiples of $15$: $15,30$ → $2$ numbers, $P(A\cap B)=\frac{2}{30}$.

Yeh overlap karte hain (jaise $15$), toh general rule use karo: $P(A\cup B)=\frac{10}{30}+\frac{6}{30}-\frac{2}{30}=\frac{14}{30}=\frac{7}{15}.$ "Neither" complement hai: $P(\text{neither})=1-\frac{14}{30}=\frac{16}{30}=\frac{8}{15}.$

Chhai faces mutually exclusive aur exhaustive hain, toh unki probabilities $1$ mein sum hoti hain: ${\sum }_{k=1}^{6}ck=c(1+2+3+4+5+6)=21c=1\Rightarrow c=\frac{1}{21}.$ Primes $\{2,3,5\}$ hain — alag faces, toh mutually exclusive; unki probabilities add karo: $P(\text{prime})=\frac{2}{21}+\frac{3}{21}+\frac{5}{21}=\frac{10}{21}.$