A sample space \(\Omega\) is the set of all possible outcomes of an experiment.
๐ Beginner's Breakdown:
- Keyword: Experiment: The action or process you are observing (e.g., tossing a coin, rolling a die).
- Keyword: Sample Space (\(\Omega\)): Think of this as the "master menu." It is a mathematical set containing every single valid outcome that could possibly happen. Nothing outside this set can occur.
๐ก Hinglish Concept:
Maan lo tum ek naya game khel rahe ho. Tumhare dimaag mein sabse pehla sawal aayega: "Is game mein result kya-kya aa sakta hai?"
- Experiment: Tumhara game khelna ya task karna.
- Sample Space (\(\Omega\)): Un saari possible possibilities ka ek box. Jo bhi hoga, isi box ke andar se niklega!
- Toss a coin: \(\Omega=\{H,T\}\).
- Roll a die: \(\Omega=\{1,2,3,4,5,6\}\).
- Toss two coins: \(\Omega=\{(H,H),(H,T),(T,H),(T,T)\}\).
- Room allocation Problem: Suppose 10 students need to be allocated in 10 rooms. \[ \Omega=\{(w_{1},\dots,w_{10}) \mid w_{i} \in \{1,\dots,10\}, w_i \neq w_j \text{ for } i \neq j\} \] \[ |\Omega| = \text{total possibilities} = 10 \times 9 \times \dots \times 1 = 10! \]
- Coupon collecting problem without replacement: There are coupons in a bag, and coupons are collected one by one without replacement.
๐ Beginner's Breakdown:
- Tossing two coins: Notice that \((H,T)\) and \((T,H)\) are treated as two different outcomes. This is because we assume the coins have separate identities (Coin 1 and Coin 2).
- Formula: Factorial (10!): When allocating 10 students to 10 rooms, the first student has 10 choices. The second has 9 left, the third has 8, and so on. Multiplying these choices (\(10 \times 9 \times 8 \dots\)) is written as \(10!\).
- Keyword: Without Replacement: Once you take an item out, you DO NOT put it back. The total number of items left decreases by 1 each time.
๐ก Hinglish Concept:
- 2 Coins: Jab do coin toss hote hain, toh \((H,T)\) aur \((T,H)\) ko alag-alag maana jaata hai, kyunki Coin 1 aur Coin 2 ki apni alag identity hai.
- 10! (Factorial) Logic: 10 dost hostel aaye hain. Pehle ke paas 10 rooms ke options hain, dusre ke paas 9 bache, teesre ke paas 8. Isiliye hum inhe multiply karte hain: \(10 \times 9 \times 8 \dots = 10!\).
- Without Replacement: Taash ke patto ki tarah. Ek baar card bahar nikal liya, toh bag mein options kam ho jayenge.
Random placement of \(r\) balls (distinguishable/labelled) in \(n\) urns/cells (distinguishable/labelled).
Example: \(r=2\), \(n=2\)
\[ |\Omega| = n^r \] \[ \Omega = \{(w_{1},w_{2}) \mid w_{1},w_{2} \in \{1,2\}\} = \{(1,2),(2,1),(1,1),(2,2)\} \]In general:
\[ \Omega = \{(w_{1},\dots,w_{r}) \mid w_{1},\dots,w_{r} \in \{1,2,\dots,n\}\} \]Note:
- \(w_{1}=4\) means the 1st ball is placed in the 4th cell.
- \(w_{2}=1\) means the 2nd ball is placed in the 1st cell.
- Thus, \(w_{i}\) denotes the position of the \(i\)-th ball.
๐ Beginner's Breakdown:
- Keyword: Distinguishable / Labelled: This means every ball and every cell is unique. Ball 1 is different from Ball 2.
- Formula: \(n^r\): Why is the total size \(n^r\)? Because Ball 1 has \(n\) choices (it can go into any cell). Ball 2 also has \(n\) choices. If we have \(r\) balls, we multiply \(n\) by itself \(r\) times: \(n \times n \times \dots \times n = n^r\).
- Reading the Notation: \((w_1, w_2) = (1, 2)\) means Ball 1 goes to Cell 1, and Ball 2 goes to Cell 2. This is fundamentally different from \((2, 1)\).
๐ก Hinglish Concept:
Urn-ball problem ko ek Gift Distribution ki tarah samjho:
- Tumhare paas \(r\) alag-alag gifts (labelled balls) hain aur \(n\) bache (labelled cells) hain.
- Kyunki kisi bhi bache ko kitne bhi gifts mil sakte hain, Pehle gift ko dene ke \(n\) tarike, dusre ko dene ke bhi \(n\) tarike. Total = \(n^r\).
- Result \((1,2)\) ka matlab hai: Pehla gift Bache #1 ko gaya, dusra Bache #2 ko. Aur \((2,1)\) uski puri ulti kahani hai.
- Toss a coin: \(r=1\), \(n=2\).
- Roll two dice: \(r=2\), \(n=6\).
- Birthdays: Possible configuration of the birthdays of 100 students. In this case, \(r=100\), \(n=365\).
- Random digits: The possible number of PIN (4-digit) numbers. \(n=10\), \(r=4\).
- Coupon Collecting: There are 10 coupons in a bag and choose three coupons with replacement. In this case: \(r=3\), \(n=10\).
๐ Beginner's Breakdown:
The Urn-Ball model acts like a master template for many real-world problems:
- Birthdays (\(365^{100}\)): Think of the 100 students as balls (\(r\)) and the 365 days of the year as the cells (\(n\)). Every student must fall into one day.
- PIN Codes (\(10^4\)): The 4 empty slots in a password are the balls (\(r\)), and the digits 0-9 are the 10 cells (\(n\)) they can choose from.
๐ก Hinglish Concept:
Ye Urn-Ball master key hai jo har jagah fit hoti hai!
- Birthdays: Class mein 100 bache (Balls) hain aur saal ke 365 din (Boxes) hain. Har bacha 365 days mein se kuch bhi choose kar sakta hai. Total outcomes = \(365^{100}\).
- ATM PIN: 4 blank jagah (Balls) hain jahan 0-9 tak 10 digits (Boxes) aa sakte hain. Isliye combinations \(10^4 = 10,000\) bante hain.
Placement of \(r\) indistinguishable/unlabelled balls in \(n\) labelled cells/bins. Find the number of distinguishable configurations.
Example: \(r=2\), \(n=2\)
\[ \Omega = \{(w_{1},w_{2}) \mid w_{1}+w_{2}=2,\; w_{1},w_{2} \ge 0\} = \{(1,1),(2,0),(0,2)\} \]In general:
\[ \Omega = \{(w_{1},\dots,w_{n}) \mid w_{1}+\dots+w_{n}=r,\; w_{1},\dots,w_{n} \ge 0\} \]Note: \(w_{i}\) denotes the number of balls in the \(i\)-th cell.
๐ Beginner's Breakdown:
- Keyword: Indistinguishable / Unlabelled: The balls look exactly identical. Because we can't tell them apart, we no longer care which ball goes into which cell. We only care about how many balls end up in each cell.
- Formula Logic: The equation \(w_{1}+\dots+w_{n}=r\) means the total count of balls across all cells must sum up to \(r\).
- Math Technique: To find the number of ways to solve this equation, we use the "Stars and Bars" combinatorics method: \(\binom{r+n-1}{n-1}\).
๐ก Hinglish Concept:
Kya ho agar balls ekdum identical hon? Ek jaise dikhne wale judwa!
- Socho tumhare paas 2 Dairy Milk chocolates (identical) hain aur 2 bache hain. Tum ye track nahi rakh sakte ki "kaunsi" chocolate kisse mili.
- Tum bas ye dekhoge ki: Pehle bache ko 2 mili (2,0), ya dono ko 1-1 mili (1,1), ya dusre bache ko dono mil gayi (0,2).
- Maths mein ise equation \(w_1 + w_2 = r\) se dikhate hain aur ise ginne ke method ko "Stars and Bars" kehte hain.
A subset of a sample space is known as an event.
Examples:
- Toss a coin: \(\Omega=\{H,T\}\), \(E=\{H\}\).
- Roll a die: \(\Omega=\{1,2,\dots,6\}\), \(E=\{2,4,6\}\).
- Urn-ball Problem (r balls in n cells): \(E = \text{1st cell is empty}\).
- \(r\)-labelled balls, \(n\)-labelled cells: \[ E=\{(w_{1},\dots,w_{r}) \mid w_{1},\dots,w_{r} \in \{2,\dots,n\}\} \]
- \(r\)-indistinguishable balls, \(n\)-labelled cells: \[ E=\{(w_{1},\dots,w_{n}) \mid w_{1}=0, \; w_{1}+\dots+w_{n}=r, \; w_{i} \geq 0\} \]
๐ Beginner's Breakdown:
- Keyword: Event (E): An event is the specific outcome (or set of outcomes) you are interested in looking for.
- Keyword: Subset: A smaller group extracted from the main Sample Space (\(\Omega\)).
- Urn-Ball Math Logic: If the event is "1st cell is empty", the math formula literally forces the balls to avoid cell 1.
- For labelled balls, they can only choose from cells \(2\) to \(n\).
- For unlabelled balls, the variable \(w_1\) (representing the count in cell 1) is forced to be exactly \(0\).
๐ก Hinglish Concept:
- Event kya hai? Agar Sample Space ek puri 3-hour ki movie hai, toh Event tumhara wo favorite action scene hai jiska tum wait kar rahe ho.
- Jaise Ludo mein: Total numbers \(\{1,2,3,4,5,6\}\) hain. Par agar tumhe token kholne ke liye sirf Even numbers chahiye toh tumhara Event \(E = \{2,4,6\}\) ban jayega.
- Formula mein agar "1st Box khali chahiye", iska sidha matlab hai ki saari balls bache hue \((n-1)\) boxes ke paas jayengi aur Box 1 ko zero balls milengi.
A sample space \(\Omega\) is the collection of all possible outcomes, while \(F\) is the collection of events under consideration.
Example: If \(|\Omega| < \infty\), then we may take \(F=\mathcal P(\Omega)\) (the power set of \(\Omega\)).
A map \(P: F \rightarrow [0,1]\) is said to be a probability measure on \(\Omega\) if the following hold:
- \(P(\Omega)=1\), \(P(\emptyset)=0\).
- If \(E_{1}, E_{2}, \dots\) is a sequence of mutually disjoint events (i.e. \(E_{i} \cap E_{j}=\emptyset\) for \(i \neq j\)), then \[ P\left(\bigcup_{i=1}^{\infty}E_{i}\right)=\sum_{i=1}^{\infty}P(E_{i}) \]
Then \(P(E)\) is said to be the probability of event \(E\).
๐ Beginner's Breakdown:
- Keyword: Probability Measure (\(P\)): A mathematical function that assigns a "score" or "weight" between 0 (0%) and 1 (100%) to an event.
- Formula: \(P(\Omega)=1\): The probability of something happening from the sample space is 100%.
- Keyword: Mutually Disjoint (\(E_{i} \cap E_{j}=\emptyset\)): This means two events cannot happen at the exact same time. For example, a coin cannot land on Heads AND Tails simultaneously.
- Rule: Additivity: If events are disjoint, you can simply add their probabilities together. (The symbol \(\bigcup\) means "Union" or "OR", and \(\sum\) means sum them up).
๐ก Hinglish Concept:
Yahan probability ko strictly napne (measure) ke rules set kiye gaye hain:
- \(P(\Omega) = 1\) ka matlab hai: "Bhaiya, kuch na kuch toh pakka hoga hi." Aur \(P(\emptyset) = 0\) matlab impossible chamatkar.
- Mutually Disjoint: Jaise India aur Australia dono ek hi world cup match nahi jeet sakte (no overlap). Agar events aapas mein takrate nahi hain, toh unki combined probability nikalne ke liye unhe chup-chap add kar do (summation).
- If \(E_{1},\dots,E_{n}\) are mutually disjoint, then \[ P\left(\bigcup_{i=1}^{n}E_{i}\right)=\sum_{i=1}^{n}P(E_{i}) \] In particular, for \(n=2\): \(P(E_{1}\cup E_{2})=P(E_{1})+P(E_{2})\).
- \(P(E^c)=1-P(E)\).
- If \(E \subseteq F\), then \(P(F \setminus E)=P(F)-P(E)\) and \(P(E) \le P(F)\).
- For any two events \(E\) and \(F\): \[ P(E \cup F) = P(E)+P(F) - P(E \cap F) \]
๐ Beginner's Breakdown:
- Formula: \(E^c\) (Complement Rule): The probability of an event NOT happening is exactly \(1\) minus the probability of it happening.
- Formula: \(E \subseteq F\) (Subset Rule): If Event E is fully inside Event F, its probability can never be larger than F's probability.
- Formula: General Addition Rule (\(P(E \cup F)\)): If events CAN happen at the same time (they overlap), you add \(P(E)\) and \(P(F)\), but you MUST subtract the overlap (\(E \cap F\)) so you don't count it twice!
๐ก Hinglish Concept:
- Complement Rule: Agar match jeetne ka chance 80% (0.8) hai, toh na jeetne (\(E^c\)) ka chance naturally \(1 - 0.8 = 0.2\) hoga.
- Subset: Ludo mein "2 aana" ye event, "Even number aane" ke event ke andar aata hai. Isliye \(E\) ki probability kabhi \(F\) se zyada nahi hogi.
- Intersection ka logic: Venn diagram socho. Jab \(P(E)\) aur \(P(F)\) ko add karte hain, toh unka beech wala overlap (\(E \cap F\)) do baar count ho jata hai. Isliye end mein ek baar subtract karna zaroori hai.
- \(\Omega=\{H,T\}\), \(P(\{H\})=\frac{2}{3}\) and \(P(\{T\})=\frac{1}{3}\) \(\implies\) \(P\) is a probability measure.
- \(\Omega=\{H,T\}\), \(P(\{H\})=p\) and \(P(\{T\})=1-p\), \(0 \le p \le 1\) \(\implies\) \(P\) is a probability measure.
- \(\Omega=\{H,T\}\). If \(P(\{H\})=\frac{1}{2}\) and \(P(\{T\})=\frac{1}{2}\), then all outcomes are equally likely.
- Roll a die: \(\Omega=\{1,2,3,4,5,6\}\). If \(P(\{1\})=\frac{2}{9}=P(\{3\})=P(\{5\})\) and \(P(\{2\})=P(\{4\})=P(\{6\})=\frac{1}{9}\), then \(P\) is a probability measure.
Let \(E = \text{outcome is even}\). Then \(E=\{2,4,6\}\). \[ P(E)=P(\{2\})+P(\{4\})+P(\{6\})=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{1}{3} \] - If all outcomes are equally likely then \[ P(E)=P(\{2\})+P(\{4\})+P(\{6\}) = \frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2} \]
๐ Beginner's Breakdown:
- Concept: Valid Probability Measures: A probability distribution doesn't have to be "fair". As long as all individual probabilities are positive and they add up exactly to \(1\), it is valid mathematically.
- Keyword: Equally Likely: This means it's a completely fair game. Every outcome has the exact same chance of happening (like a standard \(1/2\) coin or \(1/6\) die).
- Biased Example: In point 4, the odd numbers have a \(2/9\) chance, while evens have a \(1/9\) chance. Adding up the evens gives \(1/9 + 1/9 + 1/9 = 3/9 = 1/3\).
๐ก Hinglish Concept:
- Zaroori nahi ki maths mein har coin "fair" ho. "Sholay" movie jaisa biased (dhokebaaz) coin bhi ho sakta hai jahan Heads aane ka chance \(2/3\) aur Tails ka \(1/3\) ho. Total \(1\) ho raha hai na? Toh ye perfectly valid hai.
- Agar Ludo ka dice andar se heavy banaya gaya ho ki Odd numbers ke chances \(2/9\) ho jayein, toh us case mein Even number aane ki total probability sirf \(1/3\) banegi. Wahi fair dice mein ye \(1/2\) hoti.
Throughout, unless specified, we assume that all outcomes are equally likely.
If \(|\Omega| < \infty\) and all outcomes are equally likely, then \(P(\{x\})=\frac{1}{|\Omega|}\) for all \(x\).
Let \(E \subseteq \Omega\). Then \(P(E)=\frac{|\{x \mid x \in E\}|}{|\Omega|}=\frac{|E|}{|\Omega|}\).
๐ Beginner's Breakdown:
- The Classic Formula: If an experiment is fair (equally likely outcomes), the probability of an event \(E\) is simply the number of outcomes in \(E\) divided by the total number of outcomes in \(\Omega\).
- Formula Translation: \(P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\)
- Warning: You can only use this shortcut formula if the experiment is 100% fair. If it's a biased coin, this formula will give the wrong answer!
๐ก Hinglish Concept:
Jab tak alag se bola na jaye, hum maan kar chalte hain ki duniya fair hai (Equally likely outcomes). Jab duniya fair hoti hai, toh ek simple sa formula lagta hai jo hum school se padhte aa rahe hain:
\(P(E) = \text{Tumhare favour ke outcomes} / \text{Total possibilities}\)
Lekin yaad rakhna, agar experiment biased hai (jaise upar dice ka example tha), toh galti se bhi favorable/total wala formula mat lagana, answer galat aayega!
Suppose \(n\) rooms need to be allotted to \(n\) students. Find the probability that the \(k\)-th student gets the room no. 1.
Solution:
\(\Omega=\{(w_{1},\dots,w_{n}) \mid w_{i} \in \{1,\dots,n\}, w_i \neq w_j \text{ for } i \neq j\}\).
\(|\Omega|=n \times (n-1) \times \dots \times 3 \times 2 \times 1=n!\)
\(E=\{(w_{1},\dots,w_{n})\in\Omega \mid w_{k}=1\}\). As \(w_{k}=1\), we have \((n-1)\) choices for the remaining.
\(|E|=(n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1=(n-1)!\)
\[ P(E)=\frac{|E|}{|\Omega|}=\frac{(n-1)!}{n!}=\frac{1}{n} \]
๐ Beginner's Breakdown:
- Step 1 (Total Ways): Arranging \(n\) distinct students into \(n\) distinct rooms is a permutation. The formula is \(n!\) (n factorial).
- Step 2 (Favorable Ways): We want a specific student (the \(k\)-th student) to get Room 1. If we force that student into Room 1, we now only have \(n-1\) students left to arrange into \(n-1\) rooms. That leaves us with \((n-1)!\) arrangements.
- Step 3 (Probability): Divide Favorable by Total. \(\frac{(n-1)!}{n!} = \frac{(n-1)!}{n \times (n-1)!} = \frac{1}{n}\).
๐ก Hinglish Concept:
- Room Allotment ki Kahani: College mein \(n\) students aaye aur \(n\) rooms hain. Total ways jisme sabko room mil sakta hai wo \(n!\) hai.
- Sawal hai ki tumhare dost (k-th student) ko Room #1 milne ke chances kya hain? Agar us dost ko Room 1 dekar lock kar dein, toh baaki bache (n-1) students, aur unke liye (n-1) rooms. Wo log (n-1)! tariko se adjust honge.
- So, probability bani \(\frac{(n-1)!}{n!}\) jo cancel hokar \(1/n\) aati hai. Logical common sense se bhi socho, toh har student ke paas Room 1 jeetne ka equal chance (1/n) hi hona chahiye!
A drunkard man is given \(n\) keys of which only one fits his door. He tries them successively (sampling without replacement). Find the probability that he opens the door in the \(k\)-th trial.
Solution: Same as problem 1. Probability is \(\frac{1}{n}\).
๐ Beginner's Breakdown:
- Keyword: Without Replacement: Once he tries a key and it fails, he tosses it away. He doesn't try it again.
- Logic equivalence: This is the exact same math structure as the Room Allotment problem. Arranging the \(n\) keys into a trial sequence has \(n!\) combinations. Forcing the correct key to appear exactly at position \(k\) leaves the other keys to be arranged in \((n-1)!\) ways. Thus, the probability is \(1/n\).
๐ก Hinglish Concept:
- Piyakkad ki kahani: Ek aadmi nashe mein ghar aata hai. Uski jeb mein \(n\) chabiyan (keys) hain aur darwaza kholne wali sirf ek hai. Wo ek-ek karke chabi try karta hai aur fail hone par chabi fenk deta hai.
- Uske randomly k-th attempt mein darwaza khulne ki probability kya hogi? Yeh bilkul pichle sawal jaisa hai. Asali chabi ke 1st, 2nd, ya k-th slot pe aane ka chance exactly barabar hota hai. Isliye answer yahan bhi \(1/n\) hi aayega.
Suppose that each of \(n\) men at a party throws his hat into the center of the room. The hats are mixed up, and then each man randomly selects a hat. Find the probabilities of the following events.
- \(E_{1}\): 1st person selects his own hat.
- \(E_{2}\): at least one man selects his own hat.
- \(E_{3}\): none of the men selects his own hat.
Solution:
\(\Omega=\{(w_{1},\dots,w_{n}) \mid w_{i} \in \{1,\dots,n\}, w_i \neq w_j \text{ for } i \neq j\}\).
\(|\Omega|=n!\)
- (i) \(E_{1}=\{(w_{1},\dots,w_{n})\in\Omega \mid w_{1}=1\}\).
\(|E_{1}|= 1 \times (n-1) \times \dots \times 2 \times 1=(n-1)!\)
\[ P(E_{1})=\frac{|E_{1}|}{|\Omega|}=\frac{(n-1)!}{n!}=\frac{1}{n} \] - (ii) Observe that \(P(E_{3})=1-P(E_{2})\).
We calculate \(P(E_{2})\) using the inclusion-exclusion formula.
๐ Beginner's Breakdown:
- Total Space: Mixing up \(n\) hats to \(n\) men is a permutation: \(|\Omega| = n!\).
- Part (i) Logic: Just like the room allotment, if we force the 1st person to get his correct hat, the remaining \(n-1\) hats are distributed in \((n-1)!\) ways. Probability = \(1/n\).
- Part (ii) & (iii) Logic: "None" (\(E_3\)) is the exact opposite (complement) of "At least one" (\(E_2\)). Finding \(E_2\) is tricky because multiple men could get their correct hats at the same time (overlapping events). We need a special formula for overlapping unions.
๐ก Hinglish Concept:
- The Hat Problem (Party mein confusion): Ek grand party chal rahi hai. Sabne apne hats utaar kar hawa mein uchaal diye aur mix kar diye. Ab andhere mein sab ek random hat pehan lete hain. Total arrangements \(n!\) hue.
- Pehle aadmi ko khud ki hat milne ki probability \(1/n\) hai (same logic).
- Sabse complex part: "Kam se kam kisi ek ko uski sahi hat mile". Yahan ek hi waqt pe 2 log, ya 3 logo ko bhi sahi hat mil sakti hai. Ye events over-lap karte hain, isliye simple Plus (+) karne se double counting hogi. Yahan maths ka super-weapon nikalna padta hai: Inclusion-Exclusion formula!
Let \(A_{1},A_{2},\dots,A_{n}\) be \(n\) events. Then
\[ P\left(\bigcup_{i=1}^{n}A_{i}\right) = \sum_{i=1}^{n}P(A_{i}) - \sum_{i_1 < i_2}P(A_{i_1} \cap A_{i_2}) + \sum_{i_1 < i_2 < i_3}P(A_{i_1} \cap A_{i_2} \cap A_{i_3}) - \dots + (-1)^{n+1}P(A_{1} \cap A_{2} \cap \dots \cap A_{n}). \]๐ Beginner's Breakdown:
- Purpose of the Formula: When you want to find the probability of a union (\(\cup\) meaning "at least one happens"), but the events overlap, you use this formula to avoid counting the overlaps multiple times.
- The Alternating Pattern:
1. Add all single event probabilities (\(+\)).
2. Subtract all pairs where 2 events happen at once (\(-\)).
3. Add back triplets where 3 events happen at once (\(+\)).
4. Continue alternating signs until you reach the intersection of all \(n\) events.
๐ก Hinglish Concept:
- Yahan entry hoti hai Inclusion-Exclusion Formula ki. Ye counting ka ek 'balance' ka khel hai.
- Pehle sabhi single events ko add karo (+). Par rukho, jahan overlap hua unhe tumne do baar gin liya! Toh un pairs ko subtract karo (-).
- Subtract karne par 3-events wale common areas jyada subtract ho gaye, toh unhe wapas add karo (+). Ye plus-minus-plus-minus ka silsila chalta rahega taaki har combination exactly 1 baar hi count ho.
Using this formula, we proceed to calculate \(P(E_{2})\) for Problem 3.
Note that \(E_{2}=\bigcup_{i=1}^{n}A_{i}\), where \(A_{i}=\{(w_{1},\dots,w_{n})\in\Omega \mid w_{i}=i\}\), \(1 \le i \le n\).
Again, \(A_{1}=\{(w_{1},\dots,w_{n}) \mid w_{1}=1\} \implies |A_{1}|=(n-1)!\).
\(A_{1} \cap A_{2}=\{(w_{1},\dots,w_{n})\in\Omega \mid w_{1}=1, w_{2}=2\} \implies |A_{1} \cap A_{2}|=(n-2)!\).
\(\dots\)
\(A_{1} \cap \dots \cap A_{n}=\{(w_{1},\dots,w_{n})\in\Omega \mid w_{i}=i, \forall 1 \le i \le n\} \implies |A_{1} \cap \dots \cap A_{n}|=1\).
Hence,
\[ P(E_{2}) = n \cdot \frac{(n-1)!}{n!} - \binom{n}{2}\frac{(n-2)!}{n!} + \binom{n}{3}\frac{(n-3)!}{n!} - \dots + (-1)^{n+1}\frac{1}{n!} \] \[ = 1 - \frac{1}{2!} + \frac{1}{3!} - \dots + (-1)^{n+1}\frac{1}{n!} \]๐ Beginner's Breakdown:
- Formula: Combinations \(\binom{n}{k}\): How many ways can we choose pairs or triplets? To choose 2 people out of \(n\), we use \(\binom{n}{2}\). For 3 people, \(\binom{n}{3}\).
- Math Simplification: Let's look closely at the term \(\binom{n}{2}\frac{(n-2)!}{n!}\).
Because \(\binom{n}{2} = \frac{n!}{2!(n-2)!}\), when you multiply it by \(\frac{(n-2)!}{n!}\), everything cancels out except the \(2!\) in the denominator!
This massive equation brilliantly collapses into a very simple, elegant series: \(1 - \frac{1}{2!} + \frac{1}{3!} - \dots\)
๐ก Hinglish Concept:
- Agar hum fix kar dein ki kisi specific \(j\) logo ko unki sahi hat mil gayi hai, toh baaki bache hue logo ke hats randomly mix hone ke \((n-j)!\) tarike hain. Aur un \(j\) logo ko select karne ke Combinations \(\binom{n}{j}\) lagenge.
- Jab formula mein in sabko multiply karke solve karte hain, toh jaadu dekho: Sab kuch cancel ho kar seedha \(1/j!\) bach jata hai.
- Isi magic ki wajah se humein ye khubsurat alternating series milti hai. (Fun fact: Agar logo ki tadaad \(n\) bohot jyada ho, toh ye probability ~63.2% pe ruk jati hai!)
For \(n=3\):
\[ P(E_{1} \cup E_{2} \cup E_{3}) = P(E_{1})+P(E_{2})+P(E_{3}) - P(E_{1} \cap E_{2}) - P(E_{2} \cap E_{3}) - P(E_{3} \cap E_{1}) + P(E_{1} \cap E_{2} \cap E_{3}). \]For \(n=2\):
\[ P(E_{1} \cup E_{2}) = P(E_{1})+P(E_{2}) - P(E_{1} \cap E_{2}). \]Proof for \(n=2\):
By property of \(P\), we have:
\(P(E_{1} \cup E_{2}) = P(E_{1} \cap E_{2}^c) + P(E_{2} \cap E_{1}^c) + P(E_{1} \cap E_{2})\).
Again, \(P(E_{1}) = P(E_{1} \cap E_{2}^c) + P(E_{1} \cap E_{2}) \implies P(E_{1} \cap E_{2}^c) = P(E_{1}) - P(E_{1} \cap E_{2})\).
Similarly, \(P(E_{2} \cap E_{1}^c) = P(E_{2}) - P(E_{1} \cap E_{2})\).
Thus,
\(P(E_{1} \cup E_{2}) = [P(E_{1}) - P(E_{1} \cap E_{2})] + [P(E_{2}) - P(E_{1} \cap E_{2})] + P(E_{1} \cap E_{2}) = P(E_{1}) + P(E_{2}) - P(E_{1} \cap E_{2})\).
๐ Beginner's Breakdown:
- Visualizing n=2: Imagine two overlapping circles (a Venn Diagram). If you add Circle 1 and Circle 2, the overlapping middle section (\(\cap\)) gets counted twice. To fix this, you subtract the overlap once. That's the formula!
- Visualizing n=3: Add all 3 circles. Subtract all 3 pair-wise overlaps. By subtracting the pair overlaps, you accidentally completely removed the exact middle center where all 3 meet. So, you must add that center (\(E_1 \cap E_2 \cap E_3\)) back at the end.
๐ก Hinglish Concept:
- Agar hum bas 2 ya 3 events ki baat karein, toh formula ka logic aankhon se dikhta hai (Venn Diagrams!).
- 2 Events: Dono circles ke area ko jodo. Par beech wala overlapping area double count hua, usko minus (-) kar do. Done!
- 3 Events: Teeno ko add karo, phir har pair (jode) ke intersection ko subtract karo. Aisa karne se teeno circles ka jo ekdum central point tha, wo poori tarah delete ho gaya! Isliye balance banane ke liye end mein teeno ke common intersection ko wapas plus (+) kar do.
Suppose \(r\) balls are placed in \(n\) boxes. Find the probability that at least one box is empty in the following cases:
- (i) balls and boxes are labelled.
- (ii) balls are unlabelled but boxes are labelled.
Solution:
Let \(A_i\) denote the event that the \(i\)-th box is empty.
\(E = \text{at least one box is empty} = \bigcup_{i=1}^{n} A_{i}\).
Apply inclusion-exclusion:
Case (i):
In this case, \(\Omega = \{(w_{1},\dots,w_{r}) \mid w_{1},\dots,w_{r} \in [n]\}\).
\(|\Omega| = n \times n \times \dots \times n = n^{r}\).
\(A_{i} = \{(w_{1},\dots,w_{r}) \in \Omega \mid w_{1},\dots,w_{r} \in [n] \setminus \{i\}\}\).
\(|A_{i}| = (n-1)^{r} \implies P(A_{i}) = \left(\frac{n-1}{n}\right)^r = \left(1-\frac{1}{n}\right)^{r}\).
\(\implies P(A_{1} \cap A_{2}) = \left(1-\frac{2}{n}\right)^{r}\).
Putting these values in (*) we get the result.
Case (ii):
Let \(E_{1} = \text{each box is non-empty}\).
\(E = E_{1}^{c} \implies P(E) = 1 - P(E_{1})\). \quad (**)
\(\Omega = \{(w_{1},\dots,w_{n}) \mid w_{1}+\dots+w_{n}=r, \quad w_{i} \ge 0\}\).
\(|\Omega| = \binom{n+r-1}{n-1}\).
\(E_{1} = \{(w_{1},\dots,w_{n}) \mid w_{1}+\dots+w_{n}=r, \quad w_{i} \ge 1\}\).
Note, \(|E_{1}| = 0\) if \(r < n\).
Assume, \(r \ge n\). Then \(|E_{1}| = \binom{r-1}{n-1}\).
We have the result from (**).
๐ Beginner's Breakdown:
- Keyword: At least one (\(\cup\)): The moment a problem asks for "at least one", you should immediately think of either Inclusion-Exclusion OR the Complement rule (1 - "None"). We use both approaches here.
- Case (i) Labelled Math: If we force Box \(i\) to be empty, all \(r\) balls must go to the remaining \((n-1)\) boxes. Out of total \(n^r\) ways, this happens in \((n-1)^r\) ways. Divide them to get the probability. Since multiple boxes can be empty, we must use Inclusion-Exclusion.
- Case (ii) Unlabelled Math (Complement Trick): Dealing with overlapping empty boxes with Stars and Bars is hard. So, we find the opposite: "What if NO box is empty?" This means every box has \(\ge 1\) ball. We pre-place 1 ball in each of the \(n\) boxes. We are left with \(r-n\) balls to distribute using stars and bars, which gives \(\binom{r-1}{n-1}\). Subtract this probability from 1!
๐ก Hinglish Concept:
- Wapas chalte hain Urn-ball ki duniya mein. Pucha hai "Kam se kam ek box khali (empty) rah jaye" iski probability kya hai?
- Labelled case: Agar "Box 1 khali rakhna hai", toh saari \(r\) balls baaki bache \((n-1)\) boxes mein jayengi. Uski probability nikal aayi. Par overlap hoga (Box 1 aur 2 ek sath khali ho sakte hain), isliye wapas Inclusion-Exclusion formula lagana padega.
- Unlabelled case: Yahan chocolates identical hain. Smart trick lagate hain! Pucha hai "Kam se kam 1 khali", toh hum iska ulta nikalenge "Sabhi boxes bhare hue hon". Sabko bhara rakhne ke liye pehle har box mein 1-1 chocolate daal do. Phir baaki bachi chocolates distribute kardo. Jo fraction aaye, usko 1 mein se minus kardo. Problem solved!
A 5-card poker hand is said to be a Full house if it consists of 3 cards of the same denomination and 2 other cards of the same denomination (of course, different from the first denomination). Thus, a full house is three of a kind plus a pair. What is the probability that one is dealt a full house?
Solution:
The set of all cards can be written as: \(S = \{(x,y) \mid 1 \le x \le 4, \ 1 \le y \le 13\}\).
\(\Omega = \{\{w_{1},\dots,w_{5}\} \mid w_{1},\dots,w_{5} \in S\}\).
\(|\Omega| = \binom{52}{5}\).
\(E = \text{Full house}\).
\(E = \{\{w_{1},\dots,w_{5}\} \in \Omega \mid w_{1}(2)=w_{2}(2)=w_{3}(2), \ w_{4}(2)=w_{5}(2), \ w_{1}(2) \neq w_{4}(2)\}\).
(Here \(w_i(2)\) denotes the 2nd coordinate, i.e., the denomination).
Number of choices:
๐ Beginner's Breakdown:
- Keyword: Denomination: The number or face on the card (e.g., King, 5, Ace). There are 13 denominations.
- Total Combinations: We are just choosing 5 cards out of 52, order doesn't matter. So we use the combination formula: \(\binom{52}{5}\).
- Favorable Logic:
1. Pick 1 denomination for the "Three of a kind" (13 choices).
2. From its 4 suits (Spades, Hearts, Diamonds, Clubs), pick 3 cards (\(\binom{4}{3}\)).
3. Pick a DIFFERENT denomination for the "Pair" (only 12 choices left).
4. From its 4 suits, pick 2 cards (\(\binom{4}{2}\)).
Multiply them all together for the total favorable hands!
๐ก Hinglish Concept:
- Casino ka scene: Socho tum Las Vegas mein Poker khel rahe ho aur tumhe "Full House" chahiye. Matlab 3 cards ek jaise number ke (like 3 Kings) aur 2 cards kisi dusre number ke (like two 5s).
- Combinatorics se banate hain: Pehle 13 ranks (A to K) mein se 1 chuno jiske 3 patte chahiye, uske 4 suits mein se 3 nikal lo (13 * C(4,3)).
- Phir bache hue 12 ranks mein se kisi 1 ko chuno aur uske 4 suits mein se 2 nikal lo (12 * C(4,2)).
- Jab in favourable combinations ko total C(52,5) se divide karoge, toh tumhe samajh aayega ki ye combination milna kismat ki baat hai (~0.14%)!
Suppose 60% of the students in a school passed in both maths and english. 20% of them failed in maths and 30% failed in english. Find the probability that a student:
- (1) failed in maths or english.
- (2) failed in both maths and english.
Solution:
In this case, no need to write sample space & event space.
Let \(E_{1} = \text{students failed in mathematics}\).
Let \(E_{2} = \text{students failed in english}\).
Given, \(P(E_{1}) = 20\% = \frac{2}{10}\).
\(P(E_{2}) = 30\% = \frac{3}{10}\).
\(P(E_{1}^{c} \cap E_{2}^{c}) = 60\% = \frac{6}{10}\). \quad (Passed in both)
(1) \(E = \text{failed in maths or english}\).
\(E = E_{1} \cup E_{2}\).
(2) \(F = \text{failed in both maths and english}\).
\(F = E_{1} \cap E_{2}\).
As \(P(E_{1} \cup E_{2}) = P(E_{1}) + P(E_{2}) - P(E_{1} \cap E_{2})\), we have:
๐ Beginner's Breakdown:
- De Morgan's Law Logic (Part 1): The event "failed in maths OR english" (\(E_1 \cup E_2\)) is the exact opposite complement of "passed in BOTH" (\(E_1^c \cap E_2^c\)). If you didn't pass both, you must have failed at least one. Therefore, \(1 - 60\% = 40\%\).
- General Addition Rule (Part 2): We know the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). We just rearrange it to solve for the intersection (both failed): \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\). Substitute the values: \(0.2 + 0.3 - 0.4 = 0.1\).
๐ก Hinglish Concept:
- Maths aur English ka result: 60% bache Maths aur English dono mein pass hain. 20% log Maths mein fail hain aur 30% English mein fail hain.
- Agar "Dono mein pass" hone wale 60% hain, toh iska opposite scenario socho. Agar tum dono mein pass nahi hue, iska matlab tum kisi na kisi ek (Maths ya English) mein toh fail hue ho. Isliye "Failed in Maths OR English" ka answer seedha \(1 - 0.6 = 0.4\) (40%) aaya.
- "Dono mein fail" nikalne ke liye basic Venn diagram ya formula lagao: P(Maths fail) + P(Eng fail) - P(Maths OR Eng fail). Yaani \(0.2 + 0.3 - 0.4 = 0.1\) (10%).