unit 1 RVSP notes

Unit-1 Random Variable

Commutative Law
Associative Law:
- $AU(BUC)=(AUB) UC$
- $A \cap (B \cap C) = (A \cap B) \cap C$
Distributive Law
- $AU (B \cap C) = (AUB) \cap (AUC)$
- $A \cap (BUC) = (A \cap B) U (A \cap C)$
Complement
- $A \cup A^c = S$
- $S^c = \phi$
- $(A^c)^c = A$
Demorgan's Law:-
- $(A \cup B)^c = A^c \cap B^c$
- $(A \cap B)^c = A^c \cup B^c$

Probability

Probability introduced to set
- Event: An event is defined as all possible outcomes of an experiment.
- Sample Space (S): Sample Space is defined as set of all possible outcomes of an experiment.

Probability

Let s be the sample space be an event associated with random experiment then probability event A is defined as the ratio of number of favarouble Outcomes of an event to the exustive number of Cases in Sample space.
- $P(A) = \frac{N(A)}{n(s)}$
  - $N(A)$ = No of favorable outcomes of an event A
  - $n(s)$ = Exclusive number of cases in Sample space.
Ex: Consider a Die of experiment Sample Space is 6 faces indicates {1, 2, 3, 4, 5, 6}
- Event A represents even number faces.
  - $A = {1, 2, 3, 4, 5, 6}$
  - $n(s) = 6$
  - $n (A) = 3$
  - $P(A) = \frac{1}{2}$
Probability introduced to AXIOMS
- AXIOM I
 - $P(A) \geq 0$
 - The probability of an event is always non negative numbers
- AXIOM II
 - $P(s) = 1$
 - i.e The probability of sample space is always unity
 - $P(s) = \frac{n(s)}{n(s)} = 1$
- AXIOM-III
 - $0 \leq P(A) \leq 1$
 - i.e The probability is allways Lays between 0 to 1.
- AXIOM - If events A and B are matually exclusive
 - $P(A \cup B) = P(A)+ P(B)$
- AXIOM -
 - If $A1, A2, A3…. An$ are mutually exclusive events then
 - $P(A1 \cup A2 \cup A3 …. \cup An) = P(A1) + P(A2)… + P(A_n)$

Properties of probability

Theorem 1: The probability of an impossible event is zero
- Proof: Sample space (s) and impossible event ( $\phi$ ) are mutually exclusive events
  - $P(s \cup \phi) = P(s)$
  - $P(s)+ P(\phi) = P(s)$
  - $\implies P(\phi) = P(S) - P(S) = 0$
Theorem 2: If $A^c$ is the complementry event of event A
- Then $P(A^c) = 1-P(A)$
  - Proof: - A and $A^c$ are mutually exclusive events
    - $A \cup A^c = S$
    - $P(A \cup A^c) = P(S)$
    - $P(A) + P(A^c) = 1$
    - $\implies P(A^c) = 1-P(A)$
Theorem 3: If A and B are two events then
- $P(A \cup B) = P(A)+P(B)-P(A \cap B)$
  - Event A is the Union of $A \cap B^c$ and $A \cap B$ that is $A = (A \cap B^c) \cup (A \cap B)$
  - Event B is the union of $A \cap B$ and $A^c \cap B$ i.e. $B = (A \cap B) \cup (A^c \cap B)$
  - $P(A) + P(B) = P((A \cap B^c) \cup (A \cap B)) + P((A^c \cap B) \cup (A \cap B))$
  - $P(A) + P(B) = P(A \cap B^c) + P(A \cap B) + P(A^c \cap B) + P(A \cap B)$
  - $A \cap B^c, A \cap B$ are mutually exculsive event and $A^c \cap B, A \cap B$ are mutually exclusive Events
    - $P(A)+P(B) = P(A \cup B) + P(A \cap B)$
    - $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Theorem 4: If events A, B and C are
- $P(A \cup B \cup C) = P(A) + P(B) + P(C) -P(A \cap B) - P(A \cap C) -P(B \cap C) + P(A \cap B \cap C)$
  - $P(A \cup B \cup C) = P(A) + P(B \cup C) - P(A \cap (B \cup C))$
  - $P(A) + P(B) + P(C) - P(B \cap C) - [P(A \cap B) \cup(A \cap C) - P(A \cap B \cap C]$
  - $= P(A) + P(B)+P(C) - P(B \cap C) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)$
Theorem 6
- If A $\subseteq$ B (A subset B) :
  - $P(A) \leq P(B)$
    - $A \cap B = A$
    - From the figure A and $A \cap B^c$ are mutually Exclusive events
    - Then $A \cup (A \cap B^c) = B$
    - $P (A \cup (A \cap B^c)) = B$
    - $P(A) + P(A \cap B^c) = P(B)$
Theorem 7-
- If B $\subseteq$ A (B Subset A)
  - $P(B) \leq P(A)$
    - From the figure B and $A \cap B^c$ are a mutually exclusive events
    - $B \cup (A \cap B^c) = A$
    - $P(B \cup (A \cap B^c)) = P(A)$
    - $P(B)+ P(A \cap B^c) = P(A)$
Problems:
- A fair Coin is fossed four times defined the Somple space corresponding to this rondom experiment. Also give the subset Corresponding to the following events and find respective probablibes
 - (i) more heads than tails are obtained,
 - (ii) Tails occurred even number Losses
 - Sol
 - A fair coin ia fossed 4 times le number of all possible outcomes
 - n(s) = 2^n = 2^4 => 16
 - defind the sample space is equal to the set of all possible cases.
 - Sample Space = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, TTHH, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}
 - (i) Event 'A' Represents more heads than tails are obtained
 - A = {HHHT, HHHT, HHTH, HTHH, THHH }
 - n (A) = 5
 - $P(A) = \frac{n (A)}{n(s)}$
 No of favorable cases of event A
 Exclusive No of cases in sample space
 - $P(A) = \frac{5}{16}$
 - (11) Event "B" represents fails occured even number Losses.
 - B = { TTHH, THTH, HTHT, HTTH, TTHH }
 - n (B) = 4
 - $P(B) = \frac{n(B)}{n(s)} = \frac{4}{16}$
- When two dies are throughn determine the Probability that.
 - (1) A = {sum = 7 }
 - (ii) B = {8<sum≤11}
 - (iii ) c = {10 < sum}
 - (iv) Determine P(B $\cap$ C)
 - (V) Determine P (B $\cup$ C)
 - Sol Two Dies are thrown
 - $S_1 = {1, 2, 3, 4, 5, 6}$
 - $S_2 = {1, 2, 3, 4, 5, 6 }$
 - Over all sample space
 - $S=S1 \times S2$
 - $S= {1, 2, 3, 4, 5, 6} \times {1, 2, 3, 4, 5, 6 }$
 - S={
 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 (3,1) (3.2) (3,3) (3,4) (3,5) (3,6)
 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
 (5, 1) (5,2) (5,3) (5,4) (5,5) (5,6)
 (6, 1) (6,2) (6,3) (6,4) (6,5) (6;6) }
 - n(s) = 36
 - (i) A = {sum =7}
 = {(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)}
 n(A) = 6
 $P(A) = \frac{N(A)}{n(s)} = \frac{6}{36} = \frac{1}{6}$
 - (11) B = {8 <sum $\le$ 11}
 {Sum = 9,10,11}
 {(3,6) (4,5) (6,3) (5.4) (4,6) (5,5) (6,4),
 (5,6) (6,5) }
 n (B) = 9
 $P(B) = \frac{n(B)}{n(s)} = \frac{9}{36} = \frac{1}{4}$
 - C = {10< Sum}
 {Sum 11,12}
 = {(5,6) (6,5) (6,6) }.
 n(c) = 3
 $P(C) = \frac{n(c)}{n(s)} = \frac{3}{36} = \frac{1}{12}$
 - (IV) $B \cap C$
 {83}
 $\le$ 11} $\cap$ { fro<sum}
 { sum = 11}
 = {(5,6) (6,5)}
 n ( $B \cap$ C) = 2
 $P (B \cap C) = \frac{n (B \cap C)}{n(s)} = \frac{2}{36} = \frac{1}{18}$
 - (V) P (B $\cup$ C) = (P(B) + P(c) - P(B $\cap$ C)
 $= \frac{1}{4} + \frac{1}{12} - \frac{1}{18}$
 $= \frac{9+3-2}{36} = \frac{10}{36} = \frac{5}{18}$

Joint and Conditional probability

Joint probability: The joint probability of two events A and B denoted as $P(A \cap B)$ we know that
- $P(A \cap B) = P(A) + P(B) = P(A \cup B)$
Conditional probability: The conditional probability of event B, Assuming that Event A has happened denoted as P(B/A) and defined as $P(B/A) = \frac{P(A \cap B)}{P(A)}$
- Provided P(A) $\neq$ 0
- $P (A/B) = \frac{P(A \cap B)}{P(B)}$ , provided P(B) $\neq$ 0
Properties
- If any two events A and B if A $\subseteq$ B then
 - P (B/A) =1
 - Proof! A $\subseteq$ B then $A \cap B = A$
 - We know that
 - $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)} = 1$
- Any Ewo events A and B if B $\subseteq$ A then
 - $P (B/A) = \frac{P(B)}{P(A)}$
 - if B $\subseteq$ A then $B \cap A = B$
 - $P(B \cap A) = P(B)$
 - $P[B/A] = \frac{P(B \cap A)}{P(A)} = \frac{P(B)}{P(A)}$
 - 0< P(A) <1
If A and B are mutually exclusive events Explain
- If A and B are mutually exclusive events
  - $A \cap B = \phi$
  - $P(A \cap B) = P(\phi) = 0$
  - $P (B/A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{P(A)} = 0$
- If A and B are mutually exclusive events then Anc and Bnc are also mutually exclusive events Hence,
  - $P ( (A \cup B)) = P (A/C) + P(B/C)$
    - Proof: Anc and Bnc are also mutua mutually exclusive events.
      - $P ((A \cup B)) = \frac{P ((A \cup B) \cap C)}{P(C)} = \frac{P ((A \cap C) \cup (B \cap C))}{P(c)}$
      - $\frac{P(A \cap C) + P(B \cap C)}{p(c)} = \frac{P(A \cap C)}{P(C)} + \frac{P(B \cap C)}{P(C)} = P(A/C) + P(B/C)$
P(B)>P(A) then P(B/A) > P(B). P(A/B) > P(A)
- \implies \frac{P(A \cap B)}{P(B)} > P(A)
- P(A \cap B) > P(A) P(B)
- $P (B/A) = \frac{P(A \cap B)}{P(A)}$
- P (B/A) = \frac{P(A \cap B)}{P(A)} > \frac{P(A) P(B)}{P(A)} = P(B)
P(A) > P(B) then P(A/B) >P (B/A)
- P[(A/B) > P [B/A] = \frac{P(A \cap B)}{P(B)} > \frac{P(A \cap B)}{P(A)} \implies P (A \cap B) P(A) > P (A \cap B) P(B)
Total probability theorem: If $B1, B2, B3-.. BN$ are set of exclusive and mutually exclusive events Associated with random experiment A is another event. Associated with $B_n$ When n is equal to the
- n = 1,2,3,4,5… N then
 $P(A) = \sum{n=1}^{N} P(Bn) P(A/B_n)$
- Proof: Event A are Associated with along with $B1, B2, B3 …. BN$ Inner Circuit Represents event A and outer circuit Represents Event B. and $B1, B2, B3 …. BN$ are mutually exclusive events then $A \cap B1, A \cap B2, A \cap B3 …. A \cap BN$ are also mutually exclusive events.
 - $A = A \cap B1 \cup A \cap B2 \cup A \cap B3 …. \cup A \cap BN$
 - $P(A) = P(A \cap B1 \cup A \cap B2 \cup A \cap B3 …. (\cup A \cap BN) = P(A \cap B1) + P(A \cap B2) + P(A \cap B3) + P(A \cap BN)$
 - $P(A) = \sum{n=1}^{N} P(A \cap Bn)$
 - Where $P (A/Bn) = \frac{P(A \cap Bn)}{P(B_n)}$

Baye's theorem

If $B1, B2, B3-.. BN$ are set of exhust and mutually exclusive event Associated with random experiment 'A' is the another event Associated with $B_n$ is equal to the M: 0,1,2,3,4. N then
- $P [Bn/A] = \frac{ P (Bn) P (A/Bn)}{\sum{i=1}^{N} P(Bi)P(A/Bi)}$
 - $P(A/Bn) = \frac{P(A \cap Bn)}{P (B_n)}$
 - $P (Bn) = \frac{P(A \cap Bn)}{P(A/B_n)}$
 - $P (Bn/A) = \frac{P(A \cap Bn)}{P(A)}$
 - $P (A \cap Bn) = P(A) P (Bn/A)$
 - $P (A \cap Bn) = P(Bn) P(A/B_n)$
 - $P (Bn/A) = \frac{ P (Bn) P (A/B_n)}{ P(A)}$
 - Using total probability theorem
 - $P(A) = \sum{i=1}^{N} P(Bi) P(A/B_i)$
 - $P (Bn/A) = \frac{ P (Bn) P (A/Bn)}{ \sum{i=1}^{N} P(Bi)P(A/Bi)}$

Independent Events

A set of events of Independent events if the occurence any Object does not depends on occurence and occurence of others
- $P (A \cap B) = P(A) P(B)$
If A and B are two independent events then P(A/B) = P(A),
- Explain
  - $P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A) P(B)}{P(B)} = P(A)$
  - P (B/A) = P(B)
    - $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A) P(B)}{P(A)} = P(B)$
properties
- If A and B are two independent events then A and $B^c$ are also independent events
  - Proof: $P (A \cap B^c) = P(A) P(B^c)$
  - $P(A \cap B^c) = A$
  - $P(A \cap B^c) = P(A) = P(A)$
  - $P(A \cap B^c) = P(A) = P(A)$
  - $P (A \cap B^c) = P(A) - P(A \cap B)$
- If A and B are two Independent, events then $A^c$ and B are also Independent events
  - Proof :- $A^c \cap B$ and $A \cap B$ are mutually exclusive events
- If A and B are two Independent events and $A^c$ and $B^c$ are also Independent events.
  - Proof: A and $B^c$ are Independent events
Problems:
- 1) A Card is drawn from well shottled pack of 52 Cards. Find the probability that the card draw win be. (1)
- (1) Red card
- (11) Squade card
- (111) Black Queen
- (IV) King of diamond
- Sol + A ordinary pack of cards contains 52 cards divided into 4' suits.
  - ★ The Red Suit are diamond and hearts
  - ★ Black suit are squades and clubs
  - ★ Each suit contains 13 cards of the following domination.
    - 2 3, 4, 5, 6, 7, 8, 9, 10, J (Jack), Q (Queen), K (King) A (Acc) to
      - J,Q and K are face cards
  - n(s) = 52
  - (i) Red Card = Event A (1) Squade card = Event B
    black queen - Event C
    (IV) King of diamonds = Event D
  - (i) Red cards:
    - These are 26 Red Cards from which I card can be drawn in 26 casas
    - n(A) = 26
    - $P(s) = \frac{26}{52}$
  - (ii) Squade Cards:
    - There are 13 squade from which I card can be drawn in 13 casas
    - n (B) = 13
    - $P(B) = \frac{n(B)}{n(s)} = \frac{13}{52} = \frac{1}{4}$
  - (1) Black Queen:
    - There are 2 black Queens from which I can be drawn in 2 boxes
    - n(c) = 2
    - $P(C) = \frac{n(c)}{n (s)} = \frac{2}{52} = \frac{1}{26}$
  - (1) King of diamond?
    - There is only I king of diamonds
    - n(D) = 1
    - $P(D) = \frac{n(D)}{n(s)} = \frac{1}{52}$
- ②In a box There are 100 resistors having resistance and Tollerence as shown in a table. Let a resistor selected from the box and Assume each resistor as the same likely wood being choosn define there events.
  - + Event A as Draw a 47- $\Omega$ Resistor
  - + Event B as Draw a resistance with 5% tolerence.
  - * Event cas Draw a 100- $\Omega$ Resistoy
    find individual, Joint and Conditional probability.
  - Total:
    - Resistor ( $\Omega$ ) Tolerence
      - 5% 10%
    - 22 $\Omega$ 10 14 = 24
      28 16 = 44
      8 = 32
      Total 62 38 = = 100
      +P(x=6)+P(X $\le$ 7) Individual) Events!)
      Event A as Dracon 472 Resistor from the fable
      n(A) = 44 n(s) = 100
      $P(A)=\frac{n(A)}{n(s)} = \frac{44}{100} =\frac{11}{25}$ +
      +Event B as drawn a resistor with 5% tolerence.
      n(B) =62
      $P(B)=\frac{n(B)}{Loun(s)} = \frac{62}{100}$ +
      Event c has Drawn a 100-a resistor
      n(c) 32
      \n(c) = 32
      P(c)=\frac{n(c)}{n(s)} \frac{32}{100} = \frac{8}{25}+
      nt probability: AMB, BMC, CNA++nM
      AMB = 28
      $P \frac{ (AMB)}{n(s)}=\frac{28}{100}$ +
      n (B $\cap$ C) = 24
      $P=\frac{ n (B \cap C)}{n(s)}=\frac{24}{100}+P (n (B \cap C))$ +
      n (C $\cap$ A) = 0]+
      Conditional probability = P(B/A); P(A), P(B/C);
      P. (C/B) P (C/A); P(A/C)+
      P (B/A)
      $P(B/A) = \frac{P(A\cap B)}{P(A)}=\frac{28/100}{44/100}$ +
      P (A/B)
      $P(A/B) = \frac{P(A\cap B)}{P(B)}=\frac{28/100}{62/100}$ +
      p (B/C)
      $P(B/C) = \frac{P(BA\cap)}{P(C)}= \frac{24/100}{32/100}$ +
      p (CIB)
      $P(CIB) = \frac{P(B\cap C)}{P(B)}= \frac{24/100}{62/100}$ +
      p (CIA)
      $P(CIA) = \frac{P(C\cap A)}{P(A)}=\frac{0}{44/100} = 0$ p (A/C)
      $P(A/C) = \frac{P(C\cap A)}{P(C)}=\frac{}{}$ +
      Show that the chances of throwing sum 6 with+4,2 or 2 dies respectingly are 1:6:1814 (1)++
      P(x=0)+ P(x-1)+ P(X So If u dies are thrown. The favourable even £++
      is "sum of 6".
      The faverable outcomes are+
      {1113, 1131, 1311, 3111, 1122, 1212, 2121,
      22112112} +No of favourable outcomes = 10++
      Total no of possible outcomes are = 6^n++]+
      ++Probability of outcomes =No Total++n+o. of favarouble cases +P₁ =+\frac{10}{6^4}+++

Case (ii) If 3 dice are thrown favourable outcomes.The favourable outcomes are+++\114, 141, 411, 123, 321, 213, 312, 132, 231, 222 }++\No of favorable+outcomes 10+++
total no of possible outcomes 6 = +
₂ =+\frac{10}{3}+
+++ +

Case (iii) If 2 dice are thrown favorable outcomes +++
A A++
The favarouble outcomes are+
{24, 42, 15, 51, 33 }++
No of favorable outcomes = 5++
Total No of possible outcomes 6² ++ +
3 =+\frac{5}{6^2}+++
+
+\frac{10}{:}\frac{10 X 64}{:}\frac{5 }{ 6^{ = 1:6:18+++
+++ +,+++
+++++++ ++ A.. A, are mutually exclusive and exhusted + set of events As with random ex with+++ Sets of E, Events P, P with+++ Rondom joint marginal with listed The +++ Probabilibes B B, Probabi+P(A, =3+\frac P\36}(A =4+\frac {2}{36}AP=2+\frac{\A_{7}}{3++ +++++ Find (B And, P3,+++++++++++++++++, B ++=++.6++++ =P(A And, B)+ P And +P+\frac {P(A B) = +
++++ Simil B) +P+ 6++++ +
+\text {P +++
+of A-6++++++++A++++++++++++++++B+++++++++++++++++++++++++A++++++++++++++++++++++++++++++++++++++++++++++0+++++++The+a++++valid++A+=0+++++++++(9)++A++++valid+++++++++++++

Random Variables: Random variables is a function + that has assign a Real number X every +element Ses Let 's' be the Simple Space+Corresponding to Random experiment. Let 'x' is a random variable (or), Stochastic variable (or)
+Condition x should be an event of real number
+The 1) probability of x= oo and x == ০ are equal +Classification of Random variables +
Discreate+Continious+
Minced Random variables +Cumulative distributive function (CDF) +++If x is random variable of Discrete (or)+Continious then function, +If then (fe (xi) ++++for++ (+)+= for where+++++++++( =+++=++for where xi +++>=1 If >x