PROBABILITY THEORY
Presented by:
DR. NISHA ARORA
Basic Terminology
Mutually Exclusive Events
Probability
Independent Events
Conditional Probability
Addition Theorem
Multiplication Theorem
Trial/ Random Experiment
An experiment performed repeatedly essentially under the same conditions
Trial: Toss a coin 20 times
Trial: Throw a die 50 times
Event
The possible outcomes of the experiment
Event: Getting Head or Tail
Event: Rolling a 3 on a die
Event: Getting an ace
The total possible outcomes of a trail
Exhaustive Events
In a throw of a die
Number of exhaustive events = 6
H
T
H
T
H
T
In a toss of two coins
Number of exhaustive events = 4
In a draw of a playing card from the deck
The outcomes of a trail which cause the happening of a particular event.
Favorable Events
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
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}
The events are said to be equally likely
events, if none of them is expected to occur in
preference to other.
For Example
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
Equally likely Events
The events which can not occur simultaneously In a draw of a card from a deck of playing cards
Mutually Exclusive/ Disjoint
Event
A = The card drawn is a club B = The card drawn is a heart
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
Probability
P(E) = Number of favorable events
Number of exhaustive events
= m
n
»Probability can be expressed in terms of fraction, percentage, decimal or ratio.
Probability of each event is a number
between 0 and 1 inclusive i. e.,
0 ≤ P(E) ≤
1
Probability of impossible event is zero.
[Note: the converse is not necessarily true]
Probability of certain event is one.
The sum of probabilities of all possible
events is equals to one i.e.,
∑ P(E) =1
Facts
Number of exhaustive cases
= Total number of balls in the urn = = 5 + 4 = 9
Number of favorable cases
= Number of blue balls in the urn = = 4
Hence, the probability of blue ball is
P(Blue Ball) = 4/9
An urn contains 4 blue balls and 5 red balls. Find the probability that a ball chosen at random from the urn is blue.
The non-happening of event E is called complementary event EC of event E.
P(E
C) = 1 – P(E)
Complementary Event E
CIf the probability of rain is 20% or 0.2, then the probability of the complement (no rain) is 1 - 0.2 = 0.8 or 80%
Independent Event: The happening/non-happening of one event does not depend on the occurrence of other event.
Independent/ Dependent Events
Dependent Event: The events which are not independent events.
In tossing an unbiased coin event of getting a head in the 1st toss is independent of getting a head in
Example
From a bag containing three red and five blue balls. Draw two balls one by one.
Let 1st drawn ball is red and 2nd drawn
ball is blue.
If the drawn ball is replaced
P (R1) = 3/8, P(B2) = 5/8
These events are independent events.
If the drawn ball is not
replaced
P (R1) = 3/8, P(B2) = 5/7
These events are dependent events.
The probability of event A provided event B
has already happened.
P (A|B) =
Concept
If an event B has occurred, instead of S, we
consider B only.
The conditional probability of A given B will
be the ratio of that part of A which is included
in B i.e. P(A⋂B) to the probability of B.
Conditional Probability P(A|B)
) ( ) ( B P B A P
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 king
B = 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
) ( ) ( B P B A P 52 / 26 52 / 2
Concept
There are total 26 red cards out of which we have to find the probability that a king is drawn.
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
For two events A and B, probability of happening atleast one of them is
P(A⋃B) = P(A) + P(B) – P(A⋂B)
If the events A and B are mutually exclusive i.e. P(A⋂B) =0, then
P(A⋃B) = P(A) + P(B)
A⋂B A⋂B = φ
Addition Theorem
The probability that a student passes a physics test is 0.65 and a math’s test is 0.55 and the probability that he passes both tests is 0.25
What is the probability that he will pass atleast one test.
Let’s define the events
A = The student pass physics test
B = The student pass math’s test P(A) = 0.65, P(B) = 0.55
P(He pass both the test) = P(A⋂B) = 0.25 P (He passes atleast one test) = P(A⋃B)
P(A⋃B) = P(A) + P(B) - P(A⋂B) ,
= 0.65 + 0.55 - 0.25
Let’s define the events
A = 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(A⋃B)
P(A⋃B) = P(A) + P(B), (As the events are mutually exclusive)
= 3/6 + 1/6 = 4/6 =2/3
If a die is thrown. What is the
probability of an even number or a three?
For two events A and B, probability of their simultaneous happening is
P(A ⋂ B) = P(A) P(B|A), P(A) > 0
Or
P(A ⋂ B) = P(B) P(A|B), P(B) > 0
If the events A and B are independent i.e. P(A|B) = P(A) & P(B|A) = P(B), then
P(A ⋂ B) = P(A) P(B)
Let’s define the events
A = 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 not replaced)
P(Both cards are red) = P(A⋂B)
P(A⋂B) = P(A) P(B|A )
(As the events are dependent)
=
P(both cards are red) =
= P(1
stcard red) * P(2
ndcard is
red | 1
stcard was red)
= (26/52) * (25/51)
= 0.2451
Two cards are dealt, one after the other,
from a shuffled 52-card deck. What is the probability of getting two red cards ?
51 25 52 26
Lets define the events
A = Getting a head {H}, P(A) = ½ B = Getting a four {4}, P(B) = 1/6 P(head & four) = P(A⋂B)
P(A⋂B) = P(A) P(B), As the events are independent.
P(A⋂B) = =
A die and a coin are thrown.
What is the probability of a head on the coin and a four on the die ?
6 1 2 1 12 1
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