probability theory dr. nisha

7/31/2019 Probability Theory Dr. Nisha

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Presented by:

DR. NISHA ARORA

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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

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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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