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7/31/2019 Probability Theory Dr. Nisha
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Presented by:
DR. NISHA ARORA
1
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Basic Terminology Mutually Exclusive Events
Probability
Independent Events Conditional Probability
Addition Theorem
Multiplication Theorem
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An experiment performed repeatedly essentiallyunder the same conditions
Toss a coin 20 times
Throw a die 50 times
Draw a card from the deck of playing cards
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The possible outcomes of the experiment
Getting Head or Tail
Rolling a 3 on a die
Getting an ace
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The total possible outcomes of a trail
In a throw of a dieNumber of exhaustive events = 6
H
T
H
T
H
T
In a toss of two coinsNumber of exhaustive events = 4
In a draw of a playing card from the deckNumber of exhaustive events = 52
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The outcomes of a trail which cause the happening ofa particular event.
A = Getting an even number = {2, 4, 6}
Number of favorable events = 3
B = Getting a number less than 4 = {1, 2, 3}
Number of favorable events = 3.Throw of a die
Draw of a card
C = Getting a king
Number of favorable events = 4
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The set of all possible outcomes of a trail
In a toss of a coin
S = {H, T}
In a throw of a die
S = {1,2,3,4,5,6}
In a toss of two coins
S = {HH,HT,TH,TT}
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The events are said to be equally likely events,if none of them is expected to occur inpreference to other.
In a toss of an unbiased coin
P (H) = P (T) = 1/2
In a throw of a fair die
P (1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6
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The events which can not occur simultaneously
In a draw of a card from a deck of playing cards
A = The card drawn is a club
B = The card drawn is a heart
Events A and B are mutually exclusive events
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If a random experiment results in n exhaustive,
mutually exclusive and equally likely events, out of
which m are favorable to the happening of event E,
then the probability of occurrence of event E is
P(E) =Number of favorable events
Number of exhaustive events
= mn
Probability can be expressed in terms offraction, percentage, decimal or ratio.
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Probability of each event is a number between0 and 1 inclusive i. e.,
Probability of impossible event is zero.
Probability of certain event is one.
The sum of probabilities of all possible eventsis equals to one i.e.,
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Number of
= Total number of balls in the urn
= = 5 + 4 = 9
Number of
= Number of blue balls in the urn
= = 4
Hence, the probability of blue ball is
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The non-happening of event E is calledcomplementary event EC of event E.
If the probability of is 20% or0.2 then the probability of thecomplement ( ) is 1 - 0.2 =0.8 or 80%
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Thehappening/non-happening of oneevent does not depend on theoccurrence of other event.
The events which are notindependent events.
In tossing an unbiased coin event of getting a headin the 1st toss is of getting a head inthe 2nd toss.
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From a bag containing three red and fiveblue balls. Draw two balls one by one.
Let 1st drawn ball is red and 2nd drawn ballis blue.
If the drawn ball is
P (R1) = 3/8, P(B2) = 5/8
These events are .
If the drawn ball is
P (R1) = 3/8, P(B2) = 5/7
These events are .
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The probability of event A provided event B has
already happened.
P (A|B) =
If an event B has occurred, instead of S, weconsider B only.
The conditional probability of A given B will bethe ratio of that part of a which is included in Bi.e. P(AB) to the probability of B.
)(
)(
BP
BAP
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Draw a card from a deck of playing cards.
What is the probability that the card is a king
when it is a red card?A = The drawn card is a kingB = The drawn card is a red card
P (B) = P (Red card)= 26/52
And P (A B) = P (King & red card)
= 2/52
By definition, P (A|B) =
= = 1/13
)(
)(
BP
BAP
52/26
52/2
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There are total 26 red cards out of which we
have to find the probability that a king isdrawn.
Exhaustive events = Total number of red cards
= 26
Favorable events = Number of kings of red cards
= 2
Hence
P(King|Red card) = 2/26
= 1/13
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For two events A and B, probability of happeningatleast one of them is
If the events A and B are i.e.
P(AB) =0, then
A B BA
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Lets define the events
A = The student pass physics test
B = The student pass maths testP(A)= 0.65, P(B)= 0.55
P(He pass both the test) = P(AB) = 0.25
P (He passes atleast one test) = P(AB)
,
= 0.65 + 0.55 - 0.25
= 0.95
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Lets define the eventsA = Rolling an even number {2, 4, 6}
B= Rolling a three {3}
P(A)= 3/6, P(B)= 1/6
P(even number or three) = P(AB)
,
(As the events are )
= 3/6 + 1/6
= 4/6 =2/3
P(AB)
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For two events A and B, probability of theirsimultaneous happening is
Or
If the events A and B are i.e. P(A|B) = P(A)
& P(B|A) = P(B), then
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Lets define the eventsA = Getting 1st red card, B = Getting 2nd red card
P(A) = 26/52(As there are 26 red cards out of 52 playing cards)
P(B|A) = P(2nd card is red| 1st card was red)= 25/51
(As the 1st drawn card is )
P(Both cards are red) = P(AB)
(As the events are )
=
51
25
52
26
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Lets define the events
A = Getting a head {H}, P(A) =
B = Getting a four {4}, P(B) = 1/6
P(head & four) = P(AB)
,
As the events are .P(AB) =
=
6
1
2
1
12
1
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Thanks
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Queries