Lectures 4–6: Conditional Probability

Lecture 4: Conditional Probability

Definition: Conditional probability

Let $E, F \subseteq \Omega$ such that $P(F) > 0$. Then the conditional probability that $E$ occurs given that $F$ has occurred is

\[ P(E \mid F) = \frac{P(E \cap F)}{P(F)} \quad \text{for } P(F) > 0. \]

$P(E \mid F)$ is also known as the probability of $E$ given $F$.

💡 Hinglish Concept:

Conditional Probability ka seedha matlab hai: "Ek cheez ho chuki hai, ab dusri ke hone ke kya chances hain?"

Socho tumne ek dice feka. Agar main tumse puchu ki '2' aane ka chance kya hai, toh tum kahoge $1/6$.
Par agar main tumhe chupke se bata du ki "Bhai, Even number aaya hai", toh ab tumhara Sample Space ($\Omega$) chota hokar sirf $\{2, 4, 6\}$ reh gaya! (Yehi hamara event $F$ hai).
Ab '2' aane ka chance $1/3$ ho jayega. Formula mein hum yehi karte hain: Dono ke ek sath hone ke chance $P(E \cap F)$ ko nayi condition $P(F)$ se divide karte hain.

Remarks

$P(E \cap F) = P(E \mid F)P(F)$.
In the uniform outcome case: \[ P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{|E \cap F|}{|\Omega|}}{\frac{|F|}{|\Omega|}} = \frac{|E \cap F|}{|F|} \]
\begin{align*} P(E_{1} \cap E_{2} \cap E_{3}) &= P(E_{1} \mid E_{2} \cap E_{3})P(E_{2} \cap E_{3}) \\ &= P(E_{1} \mid E_{2} \cap E_{3})P(E_{2} \mid E_{3})P(E_{3}). \end{align*}

💡 Hinglish Concept:

Multiplication Rule: Agar tumhe chahiye ki $E$ aur $F$ dono hon, toh pehle socho $F$ ke hone ke kya chance hain $P(F)$, aur fir agar $F$ ho gaya, toh $E$ ke kya chance hain $P(E \mid F)$. Dono ko multiply kar do!
Chain Rule: Ye 3 ya usse zyada events ke liye domino effect jaisa hai. Pehle 3rd event hua $P(E_3)$, fir 3rd ke baad 2nd hua $P(E_2 \mid E_3)$, aur last mein dono ke baad 1st hua $P(E_1 \mid E_2 \cap E_3)$.
Equally Likely Shortcuts: Agar sab outcomes fair hain, toh probabilities ko unke raw counts $|E \cap F| / |F|$ se direct replace kar sakte hain.

Law of Total Probability

Let $E_{1},\dots,E_{n} \subseteq \Omega$ such that $\bigcup_{i=1}^{n} E_{i} = \Omega$ and $E_{i} \cap E_{j} = \emptyset$ for $i \neq j$.
Let $E \subseteq \Omega$. Then

\[ P(E) = \sum_{i=1}^{n} P(E \mid E_{i}) P(E_{i}). \]

Proof: Note $E = \bigcup_{i=1}^{n} (E \cap E_{i})$.

\begin{align*} \implies P(E) &= \sum_{i=1}^{n} P(E \cap E_{i}) \\ &= \sum_{i=1}^{n} P(E \mid E_{i}) P(E_{i}). \end{align*}

💡 Hinglish Concept:

Ye rule wahan kaam aata hai jahan ek outcome tak pahunchne ke alag-alag raste (cases) hote hain.

Maan lo tumhare paas 3 alag-alag bags hain ($E_1, E_2, E_3$) aur tumhe ek Red ball nikalni hai (Event $E$).
Toh Total Probability kaise nikaloge? Tum har bag ko chunne ki probability $P(E_i)$ ko us bag se Red ball aane ki probability $P(E \mid E_i)$ se multiply karoge, aur end mein sabko add kar doge.
Basically: "Har raste ki probability" $\times$ "Us raste pe success ki probability" = Total success.

Bayes Formula

\[ P(E_{j} \mid E) = \frac{P(E \mid E_{j})P(E_{j})}{\sum_{i=1}^{n} P(E \mid E_{i})P(E_{i})}. \]

Proof:

\[ P(E_{j} \mid E) = \frac{P(E \cap E_{j})}{P(E)} = \frac{P(E \mid E_{j})P(E_{j})}{\sum_{i=1}^{n} P(E \mid E_{i})P(E_{i})}. \]

💡 Hinglish Concept:

Agar ek bimari (Event $E$) already fail chuki hai, toh uske peeche kaunsa reason (Case $E_j$) tha? Ye ulta hisaab Bayes Formula se nikalta hai.

Formula ka pehla hissa $P(E \mid E_j)P(E_j)$ batata hai us specific reason $E_j$ ki wajah se bimari aane ka chance.
Aur denominator batata hai Total Probability (saare reasons ki wajah se bimari aane ka total chance).
Dono ko divide kar do toh pata chal jayega ki specific reason ka kya hissa (contribution) tha!

Example 1

Roll a die: $F = \text{the event that an even number occurs}$, $E = \{1,2,3\}$.
Then

\[ P(E \mid F) = \frac{|E \cap F|}{|F|} = \frac{1}{3} \]

(Since $F = \{2,4,6\}$, $E \cap F = \{2\} \implies |F|=3, |E \cap F|=1$).

💡 Hinglish Concept:

Ludo Dice Ka Jaadu: Tumne dice feka aur maine bataya "Dekho, Even number aaya hai" (Event $F$).
Jaise hi maine Even bola, tumhare dimaag se 1, 3, 5 gayab ho gaye. Ab possible numbers sirf $\{2, 4, 6\}$ bache. Toh naya denominator ban gaya 3.
Tumhe jeetne ke liye Event $E = \{1, 2, 3\}$ chahiye tha. Naye bache hue numbers mein se sirf '2' hi tumhare kaam ka hai.
Isiliye naya chance $1/3$ hoga! (Favourable = 1, Total = 3).

Example 2

A family has two children. What is the conditional probability that both are boys given that at least one of them is a boy?

Solution:
$\Omega = \{(b,b), (b,g), (g,b), (g,g)\}$.
$E = \text{both are boys} = \{(b,b)\}$.
$F = \text{at least one child is a boy} = \{(b,b), (b,g), (g,b)\}$.

\[ P(E \mid F) = \frac{|E \cap F|}{|F|} = \frac{1}{3} \]

💡 Hinglish Concept:

Ye bohot famous trick question hai!

Normal case mein 2 bacho ki age ke hisaab se 4 combinations bante hain: $BB, BG, GB, GG$. (Jahan B=Boy, G=Girl aur pehla letter bade bache ke liye hai).
Humein bataya gaya ki "Kam se kam ek Boy hai". Iska matlab $GG$ wala option toh cancel ho gaya.
Ab kitne options bache? Teen! $\{BB, BG, GB\}$.
In teen mein se dono Boys ($BB$) hone ka option sirf 1 hai. Toh answer ban gaya $1/3$.

Example 3: Room allocation

($n$ rooms are allocated to $n$ persons).
$E_{1}^{c} = \text{the room \#1 is not occupied by the first person}$.
$E_{2} = \text{the second person gets the room \#1}$.
Find $P(E_{2} \mid E_{1}^{c})$, $P(E_{2})$.

Solution:
$\Omega = \{(w_{1},\dots,w_{n}) \mid w_{i} \in [n], w_i \neq w_j\}$.
$|\Omega| = n!$
$E_{1}^{c} = \{(w_{1},\dots,w_{n}) \in \Omega \mid w_{1} \neq 1\} \implies |E_{1}^{c}| = (n-1) \times (n-1)!$
$E_{2} = \{(w_{1},\dots,w_{n}) \in \Omega \mid w_{2} = 1\} \implies |E_{2}| = (n-1)!$
As $E_{2} \subseteq E_{1}^{c}$, $E_{2} \cap E_{1}^{c} = E_{2}$.

\[ P(E_{2} \mid E_{1}^{c}) = \frac{|E_{2} \cap E_{1}^{c}|}{|E_{1}^{c}|} = \frac{|E_{2}|}{|E_{1}^{c}|} = \frac{(n-1)!}{(n-1) \times (n-1)!} = \frac{1}{n-1} \]

Now, using total probability (or multiplication rule):

\begin{align*} P(E_{2}) &= P(E_{2} \mid E_{1}^{c})P(E_{1}^{c}) + P(E_{2} \mid E_{1})P(E_{1}) \\ &= \frac{1}{n-1} \times \frac{n-1}{n} + 0 = \frac{1}{n} \end{align*}

(Since $P(E_{2} \mid E_{1}) = 0$).

💡 Hinglish Concept:

Room Allocation: Total $n$ log hain aur $n$ rooms hain.
Humein pehle se pata hai ki Pehle Aadmi (Person 1) ko Room #1 nahi mila (Event $E_1^c$). Iska matlab Room #1 abhi bhi khali pada hai.
Ab Dusre Aadmi (Person 2) ki bari aati hai. Kyunki Room #1 khali hai aur ab $n-1$ log bache hain, usko Room #1 milne ka chance seedha $1/(n-1)$ ho jata hai.
Lekin agar hume kuch nahi pata hota, toh har aadmi ka kisi bhi specific room ko paane ka chance initially $1/n$ hi hota.

Example 4

A deck of 52 playing cards, containing all 4 kings, is randomly divided into 4 sets of 13 cards each. Define:
$E_{i} = \text{the } i\text{-th set has exactly one king for } i = 1,2,3,4$.
Find $P(E_{1} \cap E_{2} \cap E_{3} \cap E_{4})$.

Solution:
From Remark we have:

\[ P(E_{1}E_{2}E_{3}E_{4}) = P(E_{1} \mid E_{2}E_{3}E_{4}) P(E_{2} \mid E_{3}E_{4}) P(E_{3} \mid E_{4}) P(E_{4}). \]

Now,

\[ P(E_{4}) = \frac{\binom{4}{1}\binom{48}{12}}{\binom{52}{13}}. \] \[ P(E_{3} \mid E_{4}) = \frac{\binom{3}{1}\binom{36}{12}}{\binom{39}{13}}. \] \[ P(E_{2} \mid E_{3}E_{4}) = \frac{\binom{2}{1}\binom{24}{12}}{\binom{26}{13}}. \] \[ P(E_{1} \mid E_{2}E_{3}E_{4}) = \frac{\binom{1}{1}\binom{12}{12}}{\binom{13}{13}} = 1. \]

Therefore,

\[ P(E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{4 \times 3 \times 2 \times 1 \times \binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}}{\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}}. \]

💡 Hinglish Concept:

Chain Rule in Action (Cards Problem): Tumhare paas 52 cards hain jinhe 4 logon mein (13-13) baantna hai. Sawal hai: Sabke paas exactly 1 King aane ka chance kya hai?
Is problem mein hume condition-pe-condition lagani padegi! Pehle hum dekhte hain chauthe aadmi ko king milne ka chance $P(E_4)$. Total 52 mein se 13 card gaye, usme 4 king mein se 1 king aur baaki 48 mein se 12 cards hone chahiye.
Fir teesre aadmi ki baari. Ab uske paas sirf 39 cards aur 3 kings bache hain! Yahi $P(E_3 \mid E_4)$ ka raaz hai.
Aise hi karte karte 2nd aur 1st aadmi ka nikalte hain. Chain rule in sab conditionals ko multiply karke final chance bata deta hai.

Example 5

Consider two urns. The first urn contains seven Red and three black balls and the second contains two red and five black balls. We flip a fair coin and then draw a ball from the first urn or the second urn depending on whether the outcome was heads or tails.

(i) Find the conditional probability that a Red ball was chosen given Head occurs.
(ii) Find the probability that a red ball was chosen.
(iii) Find the conditional probability that heads occurs given that a red ball was chosen.

Solution:
Let $E_{H} = \text{the event that heads occur}$.
$E_{T} = \text{the event that tails occur}$.
$E_{R} = \text{the event that a Red ball was chosen}$.

(i)

\[ P(E_{R} \mid E_{H}) = \frac{7}{10} \]

(Since Urn 1 has 7 Red out of 10 total).

(ii)

\begin{align*} P(E_{R}) &= P(E_{R} \cap E_{H}) + P(E_{R} \cap E_{T}) \\ &= P(E_{R} \mid E_{H})P(E_{H}) + P(E_{R} \mid E_{T})P(E_{T}) \\ &= \left(\frac{7}{10} \times \frac{1}{2}\right) + \left(\frac{2}{7} \times \frac{1}{2}\right) \\ &= \frac{7}{20} + \frac{1}{7} = \frac{49+20}{140} = \frac{69}{140}. \end{align*}

(iii) Using Bayes Formula:

\begin{align*} P(E_{H} \mid E_{R}) &= \frac{P(E_{R} \mid E_{H})P(E_{H})}{P(E_{R})} \\ &= \frac{\frac{7}{10} \times \frac{1}{2}}{\frac{69}{140}} = \frac{\frac{7}{20}}{\frac{69}{140}} \\ &= \frac{7}{20} \times \frac{140}{69} = \frac{49}{69}. \end{align*}

💡 Hinglish Concept:

Ek aur classic problem: Coin fekna aur Urn choose karna!

(i) Direct Condition: Agar mujhe pata hai Head aaya hai, toh mai seedha Urn 1 mein dekhunga jahan 10 mein se 7 balls Red hain. Toh chance $7/10$.
(ii) Total Probability: Agar Red ball nikalni hai bina kuch bataye, toh dono raste check karne padenge: "Head aane aur fir Red nikalne ka chance" PLUS "Tail aane aur fir Red nikalne ka chance". Formula isi ko add karta hai.
(iii) Bayes' Theorem (Ulta hisaab): Ab mujhe pata hai Red ball hath mein hai, par coin ka result nahi pata! Toh main Total Probability ko base (denominator) bana lunga, aur sirf "Head wale raste se Red aane" ke chance ko numerator rakhunga. Isse hume ulta rasta mil jata hai!

Lecture 5: Problems on Conditional Probability

Problem 1 (Monty Hall Problem)

Suppose we are participating in a game show. Given the choice of three doors, behind one door is a car; behind the others, goats. You are asked to pick a door, say you choose No. 1, and the host, who knows what's behind the doors, opens another door, say No. 2, which has a goat. He then says to you, "Do you want to switch to door No. 3?" Does the probability of winning the car change if we switch our choice? Why?

Solution:
Suppose your 1st choice is Door 1 (D1).
$A = \text{event that the car is behind the door \#1}$.
$B = \text{event that host opens the door \#2}$.
$P(A \mid B) = ?$
$A^c = \text{car is not behind door \#1 (so it's behind D2 or D3)}$.
A goat is behind door \#2 (D2) and $P(A^c \mid B) = ?$

$\Omega = \{(D1,D2), (D1,D3), (D2,D3), (D3,D2)\}$.
Here, $\omega = (\omega_{1},\omega_{2}) \in \Omega$
$\omega_{1}$ denotes the position of the car.
$\omega_{2}$ denotes the door that Host opens.
Your 1st choice is D1.

Position of the car	Host open	Probability
D1	D2	1/6
D1	D3	1/6
D2	D3	1/3
D3	D2	1/3

\begin{align*} P(A \cap B) &= P(\{(D1,D2)\}) = \frac{1}{6} \\ P(A^c \cap B) &= P(\{(D2,D2)\}) + P(\{(D3,D2)\}) = 0 + \frac{1}{3} \end{align*} \begin{align*} P(B) &= P(A \cap B) + P(A^c \cap B) \\ &= \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{1}{2} \end{align*} \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/6}{1/2} = \frac{1}{3} \] \[ \implies P(A^c \mid B) = \frac{P(A^c \cap B)}{P(B)} = \frac{1/3}{1/2} = \frac{2}{3} \]

💡 Hinglish Concept:

Monty Hall Problem: Ye duniya ki sabse famous mind-bending problems mein se ek hai!
Tumne Door 1 choose kiya (3 doors hain, 1 ke piche Car hai). Host, jisko sab pata hai, ek aur door khol deta hai jiske piche goat hai. Ab wo puchta hai: "Door switch karna hai?"
Logical lagta hai ki ab 2 doors bache toh chance $50-50$ hoga, haina? Nahi!
Agar tum switch karte ho toh chance badh kar $2/3$ ho jata hai! Kyunki tumhari original door sahi hone ka chance sirf $1/3$ tha, aur switch karne pe bacha hua pura $2/3$ chance tumhe mil jata hai (Host ne to waste wala door waise bhi hata diya). Bayes theorem bhi hume wahi mathematically dikhata hai.

Problem 2

Suppose there are $n$ types of coupons and that each coupon collected is with replacement. Suppose $k$ coupons are to be collected. If $A_i$ is the event that there is at least one type $i$ coupon among those collected. Then find:

(a) $P(A_{1})$
(b) $P(A_{1} \cup A_{2})$
(c) $P(A_{1} \mid A_{2})$

Solution:
$E_l = \text{event that coupon type 1 is collected in } l\text{-th trial}$.

\[ A_{1} = \bigcup_{l=1}^{k} E_{l} \]

(a)

\begin{align*} P(A_{1}) &= 1 - P(A_{1}^c) \\ &= 1 - P\left(\bigcap_{l=1}^{k} E_{l}^c\right) \\ &= 1 - \left(1-\frac{1}{n}\right)^k. \end{align*}

Because $\left|\bigcap_{l=1}^{k} E_{l}^c\right| = (n-1)^k$ and $|\Omega| = n^k$, so $P\left(\bigcap_{l=1}^{k} E_{l}^c\right) = \left(1-\frac{1}{n}\right)^k$.

(b)

\begin{align*} P(A_{1} \cup A_{2}) &= 1 - P(\text{neither type 1 nor type 2 coupons are collected}) \\ &= 1 - \left(1-\frac{2}{n}\right)^k \quad \text{(Similar to a)} \end{align*}

(c)

\[ P(A_{1} \mid A_{2}) = \frac{P(A_{1} \cap A_{2})}{P(A_{2})}. \]

We have, $P(A_{1} \cup A_{2}) = P(A_{1}) + P(A_{2}) - P(A_{1} \cap A_{2})$.

\[ 1 - \left(1-\frac{2}{n}\right)^k = \left[1 - \left(1-\frac{1}{n}\right)^k\right] + \left[1 - \left(1-\frac{1}{n}\right)^k\right] - P(A_{1} \cap A_{2}) \]

Thus,

\[ P(A_{1} \cap A_{2}) = 1 - 2\left(1-\frac{1}{n}\right)^k + \left(1-\frac{2}{n}\right)^k \] \[ P(A_{1} \mid A_{2}) = \frac{1 - 2\left(1-\frac{1}{n}\right)^k + \left(1-\frac{2}{n}\right)^k}{1 - \left(1-\frac{1}{n}\right)^k}. \]

💡 Hinglish Concept:

Coupons Collect Karna: Socho chips ke packet mein alag-alag toy aate hain (Total $n$ types). Tumne lagatar $k$ packets kharide.
(a) Type 1 ka toy na aane ka chance har packet mein $1 - 1/n$ hai. Toh $k$ packets mein lagaatar na aane ka chance $(1 - 1/n)^k$ hua. Hamein chahiye ki kam se kam ek baar toh aaye! Isliye hum ise 1 se subtract (complement) kar dete hain.
(b) Type 1 aur 2 dono bilkul na aane ka chance har packet mein $1 - 2/n$ hai. Isko bhi 1 se subtract kiya jata hai.
(c) Yahan pecheeda mamla hai: Hume dono Toys ke ek sath milne ke chances nikalne hote hain (Intersection). Jiske liye hum Inclusion-Exclusion formula lagate hain, aur fir conditional formula mein daalte hain.

Problem 3

An insurance company believes that people can be divided into two distinct classes: those who are accident-prone and those who are not. During any given year, an accident-prone person will have an accident with probability 0.4 whereas the corresponding figure for a person who is not prone to accident is 0.2. If we assume that 30 percent of the population is accident-prone.

(1) What is the probability that a new policyholder will have an accident within a year of purchasing a policy?
(2) What is the conditional probability that a new policyholder will have an accident in his/her 2nd year of policy ownership, given that the policyholder has had an accident in the 1st year?

Solution:
Let $A = \text{the policyholder is accident-prone}$.
$A_i = \text{he or she has had an accident in } i\text{-th year}$.
Need to find: (1) $P(A_1) = ?$ \quad (2) $P(A_2 \mid A_1) = ?$

(1) Using Total Probability Theorem:

\begin{align*} P(A_1) &= P(A_1 \cap A) + P(A_1 \cap A^c) \\ &= P(A_1 \mid A)P(A) + P(A_1 \mid A^c)P(A^c) \\ &= 0.4 \times \frac{3}{10} + 0.2 \times \frac{7}{10} \\ &= 0.12 + 0.14 = 0.26. \end{align*}

(2)

\[ P(A_2 \mid A_1) = \frac{P(A_1 \cap A_2)}{P(A_1)} \] \begin{align*} P(A_1 \cap A_2) &= P(A_2 \cap A_1 \cap A) + P(A_2 \cap A_1 \cap A^c) \\ &= P(A_2 \mid A \cap A_1)P(A_1 \mid A)P(A) + P(A_2 \mid A^c \cap A_1)P(A_1 \mid A^c)P(A^c) \\ &= 0.4 \times 0.4 \times 0.3 + 0.2 \times 0.2 \times 0.7 \end{align*}

(Assuming accidents in different years are conditionally independent given the type of person).

\[ P(A_2 \mid A_1) = \frac{0.4 \times 0.4 \times 0.3 + 0.2 \times 0.2 \times 0.7}{0.26} = \frac{0.048 + 0.028}{0.26} = \frac{0.076}{0.26} = \frac{76}{260} = \frac{19}{65}. \]

💡 Hinglish Concept:

Accident Prone Logic: Kuch log bohot laparwah hote hain (Accident-prone), jinka accident ka chance 40% (0.4) hai. Baaki logo ka sirf 20% (0.2) hai.
(1) Pehle saal accident ka kya chance hai? Hame ye nahi pata ki naya banda laparwah hai ya nahi. Toh hum Total Probability lagayenge: (Laparwah hone ka chance $\times$ uska accident chance) + (Normal hone ka chance $\times$ uska accident chance). Yahi hume $0.26$ deta hai.
(2) Ye mazedar hai! Agar bande ne pehle saal accident kar diya, toh ek hint mil gaya ki shayad wo laparwah type (0.4) wala banda hai! Isliye dusre saal uska accident chance ab normal $0.26$ se badh kar lagbhag $0.29$ (yani $19/65$) ho jayega.

Problem 4

A bin contains 3 types of disposable flashlights. The probability that a type 1 flashlight will give more than 100 hours of use is 0.7, with corresponding probabilities for type 2 and type 3 flashlights being 0.4 and 0.3 respectively. Suppose that 20% of the flashlights in the bin are type 1, 30% are type 2 and 50% are type 3.

(1) What is the probability that a randomly chosen flashlight will give more than 100 hrs of use?
(2) Given that a flashlight lasted more than 100 hrs, what is the conditional probability that it was type 1?

Solution:
$A = \text{flashlight lasted more than 100 hours}$.
$F_i = \text{flashlight of type } i$.

(1)

\begin{align*} P(A) &= P(A \mid F_1)P(F_1) + P(A \mid F_2)P(F_2) + P(A \mid F_3)P(F_3) \\ &= 0.7 \times 0.2 + 0.4 \times 0.3 + 0.3 \times 0.5 = 0.41 \end{align*}

(2) Using Bayes Theorem:

\[ P(F_1 \mid A) = \frac{P(A \mid F_1)P(F_1)}{P(A)} = \frac{0.7 \times 0.2}{0.41} = \frac{0.14}{0.41} = \frac{14}{41} \]

Hinglish anuvaad

Bin ki flashlights type 1, 2 aur 3 ki hain: proportions 20%, 30%, 50%. Unke 100 hours se zyada chalne ke chances 0.7, 0.4, 0.3 hain. Random flashlight chalne aur chalne ke baad type 1 hone ke chances poochhe hain.

📌 Formula: Total probability + Bayes

\[P(A)=0.41,\qquad P(F_1\mid A)=\frac{0.14}{0.41}=\frac{14}{41}\]

English content ka matlab, step by step

Teen types alag cases hain aur proportions 0.2+0.3+0.5=1. Type-wise success contributions 0.7×0.2=0.14, 0.4×0.3=0.12, 0.3×0.5=0.15; total 0.41.

Success already observed ho to Bayes mein type 1 ka contribution 0.14 numerator, saari successful flashlights 0.41 denominator; answer 14/41.

Reference: For more problems of this kind see "A First Course in Probability" - Sheldon Ross.

Hinglish anuvaad

Yahaan further practice ke liye Sheldon Ross ki probability book ka reference diya gaya hai. Agla section independence ki definition shuru karta hai.

English content ka matlab, step by step

Original reference aur uske baad aane wale independence section ka link: conditional probability P(E|F) information F ke baad E ka chance hai. Agle section mein check karenge ki F ki information E ka chance badalti bhi hai ya nahi.

Lecture 6: Independent Events

Definition: Independence

Two events $E$ and $F$ are said to be independent if

\[ P(E \cap F) = P(E)P(F). \]

We say $E$ and $F$ are dependent if they are not independent.

💡 Hinglish Concept:

Independent Events kya hain? Agar pehle event (jaise barish hona) ka dusre event (jaise coin toss mein Heads aana) par koi farq nahi padta, toh wo independent hain.
Mathematically, agar $P(E \cap F) = P(E)P(F)$ hai, tabhi unko independent manenge.
Important Note: 'Independent' aur 'Mutually Exclusive' alag hote hain! Mutually exclusive matlab ek hoga toh dusra ho hi nahi sakta (unka intersection 0 hota hai). Jabki independent mein dono ek sath ho sakte hain, bas ek dusre par asar nahi daalte.

Remarks

Suppose $E$ and $F$ are independent then

\[ P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{P(E)P(F)}{P(F)} = P(E). \]

(1) If $E$ and $F$ are independent then $E$ and $F^c$ are also independent.
Proof:
\begin{align*} P(E \cap F^c) &= P(E) - P(E \cap F) \\ &= P(E) - P(E)P(F) \\ &= P(E)(1 - P(F)) \\ &= P(E)P(F^c). \end{align*} Hence $E$ and $F^c$ are independent.
(2) (H.W.) If $E$ and $F$ are independent then $E^c$ and $F^c$ are also independent.

💡 Hinglish Concept:

Agar main tumse kahu ki "Aaj barish hone" ($E$) ka "Tumhare coin mein Head aane" ($F$) se koi lena-dena nahi hai (Independent).
Toh iska matlab ye bhi hai ki "Aaj barish hone" ($E$) ka "Tumhare coin mein Tails aane" ($F^c$) se bhi koi lena-dena nahi hoga!
Isi wajah se agar $E$ aur $F$ independent hain, toh unke complements bhi aapas mein hamesha independent hi rehte hain.

Example 1: Coin toss twice

$\Omega = \{(H,H), (H,T), (T,H), (T,T)\}$.
Let $E_1 = \text{the outcome of the first toss is } H$.
Let $E_2 = \text{the outcome of the 2nd toss is } H$.

(a) Suppose all outcomes are equally likely. Then $E_1$ and $E_2$ are independent.
Proof:
Note $E_1 = \{(H,T), (H,H)\} \implies P(E_1) = \frac{1}{2}$
$E_2 = \{(T,H), (H,H)\} \implies P(E_2) = \frac{1}{2}$
$P(E_1 \cap E_2) = P(\{(H,H)\}) = \frac{1}{4} = P(E_1) \times P(E_2)$.
Hence, independent.

(b) Suppose $P(\{(H,H)\}) = p^2$, $P(\{(H,T)\}) = P(\{(T,H)\}) = pq$, $P(\{(T,T)\}) = q^2$.
For $0 < p < 1$ & $p+q=1$.
Then $E_1$ and $E_2$ are independent (H.W.).

(c) Suppose $P(\{(H,H)\}) = \frac{1}{3}$, $P(\{(H,T)\}) = P(\{(T,H)\}) = P(\{(T,T)\}) = \frac{2}{9}$.
Then $E_1$ and $E_2$ are not independent.
In this case $P(E_1) = \frac{1}{3} + \frac{2}{9} = \frac{5}{9}$
$P(E_2) = \frac{5}{9}$
$P(E_1 \cap E_2) = \frac{1}{3} \neq \frac{5}{9} \times \frac{5}{9} = P(E_1)P(E_2)$
Hence $E_1$ and $E_2$ are dependent.

💡 Hinglish Concept:

Kya ek coin ka pehla toss dusre toss ko effect karta hai?

(a) Agar coin Fair hai (chance $1/2$), toh pehla toss ($E_1$) aur dusra toss ($E_2$) pakka independent hain. (Unka multiplication check pass ho jata hai).
(b) Agar coin Biased hai (jaise Head aane ka chance $p$), tab bhi dono tosses independent hi rehte hain. Kyunki sika (coin) ko memory nahi hoti!
(c) Lekin agar tum kisi ajeeb magic coin se khel rahe ho jahan $P(\{H,H\}) = 1/3$ hai aur baaki sab $2/9$, toh ab pehle toss ka result dusre par asar daal raha hai (Math fail ho jata hai). Yahan wo Dependent ban jayenge.

Example 2: Room allocation problem

($n$ rooms are allocated to $n$ persons).
$E_1 = \text{the event that 1st person gets room \#1}$.
$E_2 = \text{the event that 2nd person gets room \#1}$.
Clearly, $E_1$ and $E_2$ are not independent. As $P(E_1) = \frac{1}{n}$, $P(E_2) = \frac{1}{n}$, but $P(E_1 \cap E_2) = 0$.

💡 Hinglish Concept:

Dependency Example: Agar 2 log ek hi Room #1 ke liye compete kar rahe hain.
Event $E_1$ hai ki "Pehle aadmi ko Room #1 mile" aur $E_2$ hai ki "Dusre aadmi ko Room #1 mile".
Kyunki Room #1 sirf ek hi aadmi ko mil sakta hai, agar pehle aadmi ko mil gaya, toh dusre ka chance seedha ZERO ho jayega.
Kyunki ek ke result ne dusre ka chance zero kar diya, ye dono strictly Dependent hain! (In fact, ye Mutually Exclusive hain).

Example 3: Roll a die twice

Assume all outcomes are equally likely.
$E_1 = \text{the sum of two outcomes is 6}$.
$E_2 = \text{the sum of two outcomes is 7}$.
$F = \text{first outcome is 4}$.
Then $E_1$ and $F$ are dependent but $E_2$ and $F$ are independent.

Solution:
(a) $E_1 = \{(w_1,w_2) \mid w_1+w_2=6, \ 1 \le w_i \le 6\} = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}$
$P(E_1) = \frac{5}{36}$
$F = \{(4,1), (4,2), \dots, (4,6)\} \implies P(F) = \frac{1}{6}$
$P(E_1 \cap F) = P(\{(4,2)\}) = \frac{1}{36} \neq \frac{5}{36} \times \frac{1}{6} = P(E_1)P(F)$.
$\implies E_1$ and $F$ are not independent.

(b) $E_2 = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$
$P(E_2) = \frac{1}{6}$
$P(E_2 \cap F) = P(\{(4,3)\}) = \frac{1}{36} = \frac{1}{6} \times \frac{1}{6} = P(E_2) \times P(F)$
$\implies E_2$ and $F$ are independent.

💡 Hinglish Concept:

Kayi baar questions mein tum bina math solve kiye sirf intuition se nahi bata sakte ki events independent hain ya dependent.
Jaise yahan: "Sum 6 aana" ($E_1$) aur "Pehli dice par 4 aana" ($F$). Inka formula test fail ho jata hai, matlab ye Dependent hain. (Agar dice par 4 aa gaya, toh Sum 6 banne ka chance badal jata hai).
Lekin "Sum 7 aana" ($E_2$) aur "Pehli dice par 4 aana" ($F$) bilkul Independent nikle! (Sum 7 aane ka chance hamesha $1/6$ hi rahega, chahe pehli dice par kuch bhi aaya ho).
Isliye humesha $P(E \cap F) = P(E)P(F)$ test laga kar check karo!

Definition: Pairwise and Mutually Independent

(1) We say $E_1, \dots, E_n$ are pairwise independent if, for any $1 \le i \neq j \le n$, \[ P(E_i \cap E_j) = P(E_i)P(E_j). \]
(2) We say $E_1, \dots, E_n$ are mutually independent if for any $2 \le r \le n$, \[ P(E_{i_1} \cap E_{i_2} \cap \dots \cap E_{i_r}) = P(E_{i_1}) \dots P(E_{i_r}), \quad 1 \le i_1 < i_2 < \dots < i_r \le n. \]

Remark: mutually independent $\implies$ pairwise independent (but converse is not true).

💡 Hinglish Concept:

Pairwise vs Mutual: Isko aise samjho:
Pairwise Independent: Jab group mein se koi bhi DO (2) log aapas mein ek dusre ko affect nahi karte.
Mutually Independent: Jab group mein kisi bhi combo (2 log, 3 log, ya sabhi) ko utha lo, aur wo ek dusre par koi effect na daalein.
Zaroori baat: Agar sab mutually independent hain, toh automatically pairwise bhi honge. Par iska ulta hamesha sach nahi hota!

Example: Pairwise but not mutually independent

$\Omega = \{1,2,3,4\}$, $P(\{i\}) = \frac{1}{4}$.
$E_1 = \{1,2\}$, $E_2 = \{1,3\}$, $E_3 = \{1,4\}$.
$P(E_k) = \frac{1}{2}$ for $k=1,2,3$.
$P(E_i \cap E_j) = P(\{1\}) = \frac{1}{4} = P(E_i)P(E_j)$ for $1 \le i \neq j \le 3$.
Hence $E_1, E_2, E_3$ are pairwise independent.
But $P(E_1 \cap E_2 \cap E_3) = P(\{1\}) = \frac{1}{4} \neq \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = P(E_1)P(E_2)P(E_3)$.
Thus, $E_1, E_2, E_3$ are pairwise independent but not mutually independent.

💡 Hinglish Concept:

Ye example prove karta ki Pairwise ka matlab Mutually nahi hota!

Yahan 3 events hain. Agar tum kisi bhi DO ko check karoge, toh wo formula paas kar lete hain (Pairwise hain).
Lekin jab tum teeno ko ek sath check karte ho $P(E_1 \cap E_2 \cap E_3)$, toh wahan formula fail ho jata hai!
Matlab do-do ke groups mein aapas mein koi dependency nahi hai, lekin teeno mil kar ek dusre par asar daal rahe hain.

Problem

Toss a coin $n$ times independently. Each time Head occurs with probability $p$. What is the probability that:

(a) at least 1 Head occurs.
(b) exactly $k$ Heads occur.
(c) all are Heads.

Solution:
Let $E_i$ be the event that Head occurs at $i$-th trial. $P(E_i) = p$.

(a) $E = \text{the event that at least 1 Head occurs} = \bigcup_{i=1}^{n} E_i$.

\begin{align*} P(E) &= 1 - P\left(\left(\bigcup_{i=1}^{n} E_i\right)^c\right) \\ &= 1 - P\left(\bigcap_{i=1}^{n} E_i^c\right) \\ &= 1 - \prod_{i=1}^{n} P(E_i^c) \quad \text{($E_i$'s are ind, so $E_i^c$'s are ind)} \\ &= 1 - (1-p)^n. \quad \text{[$P(E_i^c) = 1-p$]} \end{align*}

(b) $F = \text{the event that exactly } k \text{ Heads occur}$.
$F = \bigcup_{1 \le i_1 < \dots < i_k \le n} E_{i_1} \dots E_{i_k} E_{i_{k+1}}^c \dots E_{i_n}^c$ (a collection of disjoint events).

\begin{align*} P(F) &= \sum_{1 \le i_1 < \dots < i_k \le n} P(E_{i_1} \dots E_{i_k} E_{i_{k+1}}^c \dots E_{i_n}^c) \\ &= \sum_{1 \le i_1 < \dots < i_k \le n} P(E_{i_1}) \dots P(E_{i_k}) P(E_{i_{k+1}}^c) \dots P(E_{i_n}^c) \\ &= \binom{n}{k} p^k (1-p)^{n-k}. \end{align*}

(c) $G = \text{the event that all Heads occur} = E_1 E_2 \dots E_n$.

\[ P(G) = P(E_1 \dots E_n) = \prod_{i=1}^{n} P(E_i) = p^n. \]

💡 Hinglish Concept:

Binomial Trials: Coin ko $n$ baar fekna har baar independent hota hai.
(a) "Kam se kam 1 Head": Iska ulta kya hoga? "Sabhi Tails"! Isliye hum $1 - P(\text{all Tails})$ karte hain. (Kyunki har bar Tails aane ka chance $1-p$ hai, toh $n$ baar ke liye $(1-p)^n$).
(b) "Exactly $k$ Heads": Hume $k$ Heads ($p^k$) aur baaki bache $n-k$ Tails ($(1-p)^{n-k}$) chahiye. Aur unko arrange karne ke tarike Combinations $\binom{n}{k}$ hote hain. Yehi Binomial formula hai!
(c) "Sabhi Heads": Bas har trial ki probability multiply karte jao, toh seedha $p^n$ mil jayega.

See the logic before reading the proof

Conditioning restricts the universe