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.
- P(A) = \frac{N(A)}{n(s)}
- 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}
- Event A represents even number faces.
- 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)
- If A1, A2, A3…. An are mutually exclusive events then
- AXIOM I
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
- Proof: Sample space (s) and impossible event (\phi) are mutually exclusive events
- 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)
- Proof: - A and A^c are mutually exclusive events
- Then 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)
- 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)
- 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)
- 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)
- P(A) \leq P(B)
- If A \subseteq B (A subset 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)
- P(B) \leq P(A)
- If B \subseteq A (B Subset 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}
- 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
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
- P (B/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
- P (B/A) = \frac{P(B)}{P(A)}
- If any two events A and B if A \subseteq B then
- 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)
- Proof: Anc and Bnc are also mutua mutually exclusive events.
- P ( (A \cup B)) = P (A/C) + P(B/C)
- If A and B are mutually exclusive events
- 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)}
- n = 1,2,3,4,5… N then
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)}
- 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)
- Explain
- 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
- If A and B are two independent events then A and B^c are also 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
- 2 3, 4, 5, 6, 7, 8, 9, 10, J (Jack), Q (Queen), K (King) A (Acc) to
- 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} = 0p (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}+++
- Resistor (\Omega) Tolerence
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