What is Probability uses define types full

*
It is the branch of
Mathematics that
studies pattern of
chance
OR

*
The idea of probability is
based on observation.
Probability describes what
happens over many, many
trials.

*The chance of coin landing
on head is 50%.
*This is because coin has
two sides so there is 50%
chances that coin will land
on head and 50% chance
that coin will land on tail
Example:

*
𝑷 𝑨 =
𝑁𝑜. 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑡𝑜 𝐴
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
Where,
𝐴 = 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝐴

*
Q:The spinner is spun at once.
Determine the probability of event?
Solution:
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒 = 10
𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 =
𝑁𝑜. 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒
𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 =
1
10
𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 = 0.1

*
*Experiment:
 Very 1st and most common term used in Probability.
 It is very clear form word experiment that we are doing some act or verb.
* Random Experiment:
Any experiment whose outcome is not pre-determined and it has more than 1
outcome.
Example:
 Tossing a coin is a Random Experiment.
- Its Outcome is not pre-determined
- It has more than 1 outcome.

*
*Outcome:
A possible result form random experiment is known as outcome.
Example:
Outcome of coin flip might be head or tail.
*Sample Space:
List of all possible outcomes from random experiment or given experiment is known as
sample space.
Sample space is a set.
It is denoted by “S”

*
*Example:
If a single dice rolling experiment, The sample space will be-
𝑺 = 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔

Event
Is a subset of sample
space
Collection of outcomes
having common
characteristics
𝐸𝑣𝑒𝑛𝑡 ⊆ 𝑆𝑎𝑚𝑝𝑙𝑒 𝑆𝑝𝑎𝑐𝑒

*
𝐒 = 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔
Even No:
E = 2, 4, 6
Odd No:
F = 1, 3, 5
These are the subgroups of sample space.

*A simple event is an event with a single
outcome (only one "answer").
E.g:
*The probability of tossing a head with a
penny. (1/2)
* 𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒆𝒗𝒆𝒏𝒕 =
𝑵𝒐.𝒐𝒇 𝒕𝒊𝒎𝒆𝒔 𝒊𝒕 𝒐𝒄𝒄𝒖𝒓
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐.𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆
*In a simple event, the numerator ("number
of times it can occur") will be 1
*A compound event is the combination
of two or more simple events (with two or
more outcomes).
*E.g:
*The probability of tossing three pennies and
getting at least 2 heads. (4/8)
* 𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒆𝒗𝒆𝒏𝒕 =
𝑵𝒐.𝒐𝒇 𝒕𝒊𝒎𝒆𝒔 𝒊𝒕 𝒐𝒄𝒄𝒖𝒓
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐.𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆
* A compound event, the numerator
("number of times it can occur") will be
greater than 1

*
Example:
Two coins are tossed.
𝑺 = 𝐻𝐻, 𝑇𝐻, 𝐻𝑇, 𝑇𝑇
𝑬 = 𝐵𝑜𝑡ℎ 𝑎𝑟𝑒 𝐻𝑒𝑎𝑑 = 𝐻𝐻
𝑭 = 𝐴𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝐻𝑒𝑎𝑑 = 𝐻𝑇, 𝑇𝐻, 𝐻𝐻
Simple Event:
𝑬 = 𝑯𝑯
Outcome is simple one.
Compound Event:
𝑭 = 𝑯𝑻, 𝑻𝑯, 𝑯𝑯
This is compound event with more than one simple event.

*
*Mutually Exclusive Event:
Event that cannot occur simultaneously.
Example 1:
1. Turning left and turning right are Mutually Exclusive (you can’t do both at the same
time)
2. Tossing a coin: Heads and Tails are Mutually Exclusive
Example 2:
Randomly choosing a day from 2020.
𝑨 = 𝑫𝒂𝒚 𝒊𝒏 𝑱𝒂𝒏𝒖𝒂𝒓𝒚
𝑩 = 𝑫𝒂𝒚 𝒊𝒏 𝒇𝒆𝒃𝒓𝒂𝒖𝒓𝒚
Event A and B are disjoint and are mutually exclusive

*
*Exhaustive Event:
The total number of all possible elementary outcomes in a
random experiment is known as ‘exhaustive events’.
E.g:
If 1 dice are rolled, what is the probability to came 1, 2, 3, 4, 5, 6 ?
Solution:
𝑛 𝑆 = 6
𝑛 𝐴 = 6
𝑃 𝐴 =
𝑛 𝐴
𝑛 𝑆
=
6
6
= 1

Example 2:
A coin is tossed. Tell whether the events are exhaustive events or
not?
Where;
Event X = If Head will appear
Event y = If tail will appear
Sol:
Both events together are exhaustive events, because one will occur during
the conduct of an experiment.

*Equally likely Events:
If events have same chanced of occurrence, then they are
said to be equally likely.
E.g:
In a single toss of fair coin, the event 𝑯 𝒂𝒏𝒅 𝑻
are equally likely.

Cont..
Example 2:
Each numeral on a die is equally likely to occur when the die is tossed.
Sample space of throwing a die: { 1, 2, 3, 4, 5, 6 }
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑐ℎ𝑜𝑠𝑒𝑛 𝑛𝑢𝑚𝑒𝑟𝑎𝑙 =
𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑙 𝑜𝑐𝑐𝑢𝑟𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒
𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑛𝑢𝑚𝑒𝑟𝑎𝑙 =
1
6

Theoretical probabilities
𝑃 1 =
1
6
𝑃 2 =
1
6
𝑃 3 =
1
6
𝑃 4 =
1
6
𝑃 5 =
1
6
𝑃 6 =
1
6
Total = 1

Independent Events Dependent events
 The outcome of one event does not
affect the outcome of other event.
 If A and B are independent event then
the probability of both occurring event
is
𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴) × 𝑃 𝐵
The outcome of one event affect the
outcome of other event is known as
dependent event.
If A and B are dependent event then
probability of both occurring event is
𝑃 𝐴 𝑎𝑛𝑑 𝐵 = 𝑃 𝐴 × P 𝐵|𝐴

*
*Independent Event :
Example 1:
If a dice is thrown twice, find the probability of getting two 5’s.
Solution:
𝑃 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 5 𝑜𝑛 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡ℎ𝑟𝑜𝑤 =
𝟏
𝟔
𝑃 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 5 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡ℎ𝑟𝑜𝑤 =
𝟏
𝟔
𝑃 𝑡𝑤𝑜 5′
𝑠 =
𝟏
𝟔
×
𝟏
𝟔
=
𝟏
𝟑𝟔

*
Example 2 :
Two cards are drawn in succession with replacement from a standard deck of
cards. What is the probability that two kings are drawn?
Solution:
𝑃 𝐾1 ∩ 𝐾2 = 𝑃 𝐾1 × 𝑃 𝐾2
𝑃 𝐾1 ∩ 𝐾2 =
4
52
×
4
52
=
1
169
Example 3 :
Two marbles are drawn with replacement from a bag containing 7 blue and 3 red
marbles. What is the probability of getting a blue on the first draw and a red on the second
draw?
Solution:
𝑃 𝐵 ∩ 𝑅 = 𝑃 𝐵 × 𝑃 𝑅
𝑃 𝐵 ∩ 𝑅 =
7
10
×
3
10
=
21
100
= 0.21

*
Dependent event:
Example 1:
A purse contains four $5 bills, five $10 bills and three $20 bills. Two bills are selected
without the first selection being replaced. Find P($5, then $5)?
Solution:
There are four $ 5 bills
𝑻𝒐𝒕𝒂𝒍 𝒃𝒊𝒍𝒍𝒔 = 𝟏𝟐
𝑷 $ 𝟓 =
𝟒
𝟏𝟐
The result of the first draw affected the probability of the second draw.
There are three $5 bills left
𝑻𝒐𝒕𝒂𝒍 𝒃𝒊𝒍𝒍𝒔 𝒍𝒆𝒇𝒕 = 𝟏𝟏
𝑃 $5 𝑎𝑓𝑡𝑒𝑟 $ 5 =
𝟑
𝟏𝟏
𝑃 $ 5, 𝑡ℎ𝑒𝑛 $ 5 =
𝟒
𝟏𝟐
×
𝟑
𝟏𝟏
=
𝟏
𝟏𝟏
The probability of drawing a $5 bill and then a $ 5 𝑏𝑖𝑙𝑙 𝑖𝑠 =
1
11

*
Q : Draw two cards in succession without replacement from a standard deck.
Find the probability of a king on the first draw and a king on the second draw?
Solution:
𝑷 𝑲𝟏 ∩ 𝑲𝟐 = 𝑷 𝑲𝟏 × 𝑷 𝑲𝟐|𝑲𝟏
𝑷 𝑲𝟏 ∩ 𝑲𝟐 =
𝟒
𝟓𝟐
×
𝟑
𝟓𝟏
=
𝟏
𝟐𝟐𝟏

*
1: The probability of an event is between 0 and 1
0 ≤ 𝑃𝑟 𝐴 ≤ 1
2: The sum of the probabilities of all possible outcomes is 1
𝑺 = 1, 2, 3, 4, 5, 6
Where,
Even No: 𝑬 = 2, 4, 6
𝑷 𝑬 =
3
6
𝑷 𝑬 =
1
2
𝑷 𝑬 = 0.5

*
*Odd No:
𝑭 = 1, 3, 5
𝑷 𝑭 =
3
6
𝑷 𝑭 =
1
2
𝑷 𝑭 = 0.5
So,
0.5 + 0.5 = 1
Sum of all possible events are equal to 1

*
3: The sum of the probability of an event occurring and it not occurring is 1.
𝑃𝑟 𝐴 + 𝑃𝑟 𝑛𝑜𝑡 𝐴 = 1
Example:
A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles.
𝑟𝑒𝑑 + 𝑝𝑟 𝑛𝑜𝑡 𝑟𝑒𝑑 = 1
5
10
+ 𝑝𝑟 𝑛𝑜𝑡 𝑟𝑒𝑑 = 1

*
4 : If event B is a subset of event A, then the probability of B is less than or equal
to the probability of A
𝑃𝑟 𝐵 ≤ 𝑃𝑟 𝐴
Example:
There are 20 people in the room: 12 girls (5 with blond hair and 7 with brown hair)
and 8 boys (4 with blond hair and 4 with brown hair).
𝒑𝒓 𝒈𝒊𝒓𝒍 𝒘𝒊𝒕𝒉 𝒃𝒓𝒐𝒘𝒏 𝒉𝒂𝒊𝒓 ≤ 𝒑𝒓 𝒈𝒊𝒓𝒍𝒔
7
20
≤
12
20

*
5: Probability of an impossible event is “0”.
Example:
Obtaining a 7 on a throw of a fair number cube
𝑃 = 0
6: Probability of sure event is “1”
Example:
Obtaining a number less than 7 on a throw of a fair cube
𝑁𝑜 𝑜𝑛 𝑐𝑢𝑏𝑒 = 1, 2, 3, 4, 5, 6
𝑷 = 1

*
*Formula Of Probability:
𝑷 𝑬 =
𝑛 𝐸
𝑛 𝑆
𝒏 𝑬
𝒏 𝑺
=
𝒏𝒐.𝒐𝒇 𝒘𝒂𝒚𝒔 𝒊𝒏 𝒘𝒉𝒊𝒄𝒉 𝒆𝒗𝒆𝒏𝒕 𝒐𝒄𝒄𝒖𝒓
𝒏𝒐.𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒑𝒂𝒄𝒆
Where,
𝑷 𝑬 = 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 ℎ𝑎𝑝𝑝𝑒𝑛𝑖𝑛𝑔 𝑜𝑓 𝑎𝑛 𝑒𝑣𝑒𝑛𝑡
𝒏 𝑬 = 𝑛𝑜. 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑖𝑛 𝑤ℎ𝑖𝑐ℎ 𝑒𝑣𝑒𝑛𝑡 𝑜𝑐𝑐𝑢𝑟
𝒏 𝑺 = 𝑛𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒

*
*If one coin is tossed, find the probability of getting a head?
Solution:
𝑺 = 𝑯, 𝑻
Event “A” getting a head
𝑨 = 𝑯
𝑷 𝑨 =
𝒏 𝑨
𝒏 𝑺
=
𝟏
𝟐
= 𝟎. 𝟓
Event “B” getting a Tail
𝑩 = 𝑻
𝑷 𝑩 =
𝒏 𝑩
𝒏 𝑺
=
𝟏
𝟐
= 𝟎. 𝟓

*
*Example 2: There are 12 𝑏𝑜𝑦𝑠 & 8 𝑔𝑖𝑟𝑙𝑠 present in the class, Then total students
𝑎𝑟𝑒 20.We have to select 1 𝑏𝑜𝑦 from total students.
*Now:
 From 𝟐𝟎 𝒔𝒕𝒖𝒅𝒆𝒏𝒕𝒔, a student will be selected, So what is the probability that he
will be a boy?
*Solution:
Formula:
𝑃 𝐸 =
𝑛 𝐸
𝑛 𝑆
𝑃 𝐸 =
12
20
𝑃 𝐸 = 0.6

*
 What is the probability that selected student will be a girl?
Solution:
𝑃 𝐸 =
8
20
𝑃 𝐸 = 0.4
So, as we know sum of all probability is 1.
0.6 + 0.4 = 1

*
A die is rolled. What is the probability that dots on the top are
𝒈𝒓𝒆𝒂𝒕𝒆𝒓 𝒕𝒉𝒂𝒏 𝟒?
Solution:
𝑺 = 1, 2, 3, 4, 5, 6
𝒏 𝑺 = 6
Dots on the top are greater than 4 = 5, 6 = 𝐸
𝒏 𝑬 = 2
Formula:
𝑷 𝑬 =
𝑛 𝐸
𝑛 𝑆
𝑷 𝑬 =
2
6
𝑷 𝑬 = 0.33

• From a box containing orange flavored sweets, Bilal takes out one sweet
without looking.
Events
I. The sweet is orange – flavored
II. The sweet is lemon flavored
Sol:
𝒏 𝑺 = 𝟏
• Suppose A is the event that sweet is orange flavored.
So favorable outcome is 𝒏 𝑨 = 1
𝑷 𝑨 =
𝑛 𝐴
𝑛 𝑆
=
1
1
= 1
*

*
• Let B the event that the sweet is lemon – flavored.
Favorable outcomes is 𝒏 𝑩 = 0
𝑷 𝑩 =
𝑛 𝐵
𝑛 𝑆
=
0
1
= 0

Pakistan and India plays a cricket match. The result is
I. Pakistan wins
II. India wins
𝐎𝐮𝐭𝒄𝒐𝒎𝒆𝒔 = 𝒏 𝑺 = 𝟑
• 3 outcomes because Pakistan may win, lose or the match is
tied.
 Let A be the event that Pakistan wins
𝑛 𝐴 = 1
𝑃 𝐴 =
𝑛 𝐴
𝑛 𝑆
=
1
3
*

 Let B be the event that India does not lose. If India does not lose
then India may win, or the match gets tied.
𝑛 𝐵 = 2
𝑃 𝐵 =
𝑛 𝐵
𝑛 𝑆
=
2
3
*

There are 5 green and 3 red balls in a box, and one is taken out.
Events:
I. The ball is Green
II. The ball is Red
Sol:
Total number of balls are 8
𝐎𝐮𝐭𝐜𝐨𝐦𝐞𝐬 = 𝒏 𝑺 = 𝟖
 Let A be the event that ball is green
𝑛 𝐴 = 5
 Let B be the event that the ball is red
𝑛 𝐵 = 3
𝑃 𝐴 =
𝑛 𝐵
𝑛 𝑆
=
3
8
*

A fair coin is tossed three time and it shows event happening:
I. One tail
II. At least one head
Possible outcomes:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
𝒏 𝑺 = 𝟖
 Let A be the event that the coin shows one tail, then favorable
outcomes are
HHT, HTH, THH
𝑛 𝐴 = 3
𝑃 𝐴 =
𝑛 𝐴
𝑛 𝑆
=
3
8
*

 Let B be the event that the coin shows at least one Head, then favorable
outcomes are
HHH, HHT, HTH, THH, HTT, THT, TTH
𝑃 𝐵 =
𝑛 𝐵
𝑛 𝑆
=
7
8
*

A die is rolled. The top shows
I. 3 or 4 dots
II. Dots less than 5
Sol:
𝑷𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 = 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔
𝒏 𝑺 = 𝟔
 Let A be the event that die shows 3 & 4
𝑛 𝐴 = 2
𝑃 𝐴 =
𝑛 𝐴
𝑛 𝑆
=
2
6
=
1
3
*

 Let B the event that top of the die shows dots less than 5 then
𝑛 𝐴 = 4
𝑃 𝐴 =
𝑛 𝐴
𝑛 𝑆
=
4
6
=
2
3
*

*
 Toss two coins. Find the probability of at least one head appearing?
Solution:
At least one head is interpreted as one head or two heads.
𝑆 = 𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇
𝑛 𝑆 = 4
How many outcomes of the event “at least one head”?
𝐴𝑛𝑠𝑤𝑒𝑟 = 3 = 𝐻𝐻, 𝑇𝐻, 𝐻𝑇
𝑃 𝐸 =
𝑛 𝐸
𝑛 𝑆
=
3
4
= 0.75

*
* If a sample space is such that,
𝑛 𝑆 = 𝑁 & 𝑜𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑁 𝑒𝑞𝑢𝑎𝑙𝑙𝑦 𝑙𝑖𝑘𝑒𝑙𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑎𝑠 𝑒𝑣𝑒𝑛𝑡 𝐸 𝑜𝑐𝑐𝑢𝑟𝑠 𝑅 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒𝑛 𝑒𝑣𝑖𝑑𝑒𝑛𝑡𝑙𝑦 "E" 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑜𝑐𝑐𝑢𝑟 𝑁 − 𝑅 𝑡𝑖𝑚𝑒𝑠
*Non occurrence of event is 𝒅𝒆𝒏𝒐𝒕𝒆𝒅 𝒃𝒚 𝑬
𝑷 𝑬 =
𝒏 𝑬
𝒏 𝑺
𝒏 𝑬
𝒏 𝑺
=
𝑹
𝑵
And,
𝑷 𝑬 =
𝒏 𝑬
𝒏 𝑺
𝒏 𝑬
𝒏 𝑺
=
𝑵 − 𝑹
𝑵
𝑵 − 𝑹
𝑵
=
𝑵
𝑵
−
𝑹
𝑵
𝑵
𝑵
−
𝑹
𝑵
= 𝟏 −
𝑹
𝑵
As we know,
𝑷 𝑬 =
𝑹
𝑵
Then,
𝑷 𝑬 = 𝟏 − 𝑷 𝑬

*
*If 3 prizes for every 1000 Raffle tickets?
Solution:
𝑃 𝑁𝑜𝑡 𝑤𝑖𝑛 = 1 − 𝑃 𝑤𝑖𝑛
1 − 𝑃 𝑤𝑖𝑛 = 1 −
3
1000
1 −
3
1000
=
997
1000
997
1000
= 0.997

*
*Example 1 : The given table below shows the result of rolling a dice 100 times. Find
the probability in which odd number occurs?
*Solution:
*𝑷 𝑬 =
𝟐𝟓+𝟏𝟑
𝟏𝟎𝟎
= 𝟎. 𝟑𝟖
Event Tally Mark Frequency
1 |||| |||| |||| |||| |||| 25
2 |||| |||| ||| 13
3 |||| |||| |||| 14
4 |||| |||| |||| |||| |||| 24
5 |||| ||| 8
6 |||| |||| |||| | 16

*
*A fair coin is tossed 30 times, the result which is tabulated below. So we have answer the
question below by reading table.
Q 1: How many times “Head” appear?
Q 2: How many times “Tail” appear?
Q 3: Estimating the probability of appearance of “Head”?
Q 4: Estimating the probability of appearance of “ Tail”?
Answer:
𝑇𝑜𝑡𝑎𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒 𝑛 𝑆 = 30
1: No. of Head appears 𝑛 𝐴 = 𝟏𝟒
2: No. Of Tail appears 𝑛 𝐵 = 𝟏𝟔
Event Tally mark Frequency
Head |||| |||| |||| 14
Tail |||| |||| |||| | 16

*
3: Probability of Head:
𝑷 𝑨 =
𝒏 𝑨
𝒏 𝑺
=
𝟏𝟒
𝟑𝟎
= 𝟎. 𝟒𝟕
4: Probability of Tail:
𝑷 𝑩 =
𝒏 𝑩
𝒏 𝑺
=
𝟏𝟔
𝟑𝟎
= 𝟎. 𝟓𝟑

*
Addition Rule:
State the probability of two events is the sum of probability that ether
will happen subtract the probability that both will happen.
Formula:
1: When A & B are disjoint.
Then,
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵
2: When A & B are overlapping or 𝑩 ⊆ 𝑨
Then,
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵

*
Example 1:
There are 20 chits marked 1, 2, 3, 4,………. 20 in a bag.
Find the Probability of picking a chit, the number written on which is multiple of 4
and multiple of 7?
Solution:
𝑆 = 1, 2, 3, 4, … … . . , 20
𝑛 𝑆 = 20
1: Let the event “A” be the event of getting multiple of 4.
𝐴 = 4, 8, 12, 16, 20
𝑛 𝐴 = 5
𝑃 𝐴 =
5
20
=
1
4
= 0.25

*
2: Let event “B” be the event of getting multiple of 7.
𝐵 = 7, 14
𝑛 𝐵 = 2
𝑃 𝐵 =
2
10
=
1
5
= 0.1
As 𝑨 & 𝑩 are disjoint sets:
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵
𝑃 𝐴 ∪ 𝐵 =
1
4
+
1
10
= 0.35

 A die is thrown. Find the probability that the dots on the top are prime numbers or odd numbers?
Solution:
𝑆 = 1, 2, 3, 4, 5, 6
𝑛 = 6
Let 𝑨 = 𝑺𝒆𝒕 𝒐𝒇 𝒑𝒓𝒊𝒎𝒆 𝒏𝒖𝒎𝒃𝒆𝒓
𝐴 = 2, 3, 5
𝑛 𝐴 = 3
Let 𝑩 = 𝑺𝒆𝒕 𝒐𝒇 𝑶𝒅𝒅 𝒏𝒖𝒎𝒃𝒆𝒓
𝐵 = 1, 3, 5
𝑛 𝐵 = 3
So,
𝐴 ∩ 𝐵 = 2, 3, 5 ∩ 1, 3, 5 = 3, 5
Now,
𝑃 𝐴 =
3
6
=
1
2

*
𝑃 𝐵 =
3
6
=
1
2
And,
𝑃 𝐴 ∩ 𝐵 =
2
6
=
1
3
So,
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵
𝑃 𝐴 ∪ 𝐵 =
1
2
+
1
2
−
1
3
=
2
3

*
*Definition:
The sample space for an experiment must often be changed when some
additional information pertaining to the outcome of the experiment is received some
outcomes as being impossible which before receiving the information were believed
possible. The probability associated with such reduced sample space are called
conditional probability
It states that:
If two events A and B are defined on sample space and if probability B is not zero, then
conditional probability of an event A, given that B has occurred.
It written as:
𝑃 𝐴
𝐵 =
𝑃 𝐴 ∩ 𝐵
𝑃 𝐵
𝐵 ≠ 0

Example:
Two coins are tossed. What is the
conditional probability that two head
results given that at least one head.
Solution:
𝑺 = 𝑯𝑯, 𝑯𝑻, 𝑻𝑯, 𝑻𝑻 = 𝟒
𝑃 𝐴
𝐵 =
𝑃 𝐴 ∩ 𝐵
𝑃 𝐵
𝐴 = 𝐻𝐻 ,
𝐵 = 𝐻𝐻, 𝐻𝑇, 𝑇𝐻 ,
𝐴 ∩ 𝐵 = {𝐻𝐻}
𝑃 𝐴 ∩ 𝐵 =
𝑛 𝐴 ∩ 𝐵
𝑛 𝑆
=
1
4
𝑃 𝐵 =
3
4
𝑃 𝐴
𝐵 =
1
4
3
4
=
1
3

Given that a under-graduate student is selected at random, what is the probability that
this student is a nurse?
Solution:
Example 1:
Restricting our attention on the column representing undergrads,
We find that 90 undergrads students, 53 are nursing majors.
Therefore,
𝑃 =
𝑁
𝑈
=
53
90
= 0.58
Profes
sion
Unde
r-
grade
s
Grade
s
Total
Nursin
g
53 47 100
Engine
ers
37 13 50
90 60 150

*
Part 2 :
Given that an engineering student is selected, find the probability that
the student is a under-graduate student?
Solution:
Restricting the sample space to the 50 engineering students, 37 of the 50 are undergrads.
Therefore,
𝑃 =
𝑈
𝐸
=
37
50
= 0.74
Profession Undergrad
s
Grades Total
Nursing 53 47 100
Engineer 37 13 50
90 60 150

*
Four items are taken from random box of 12 items and inspected. The box is rejected if more than item
is found to be faulty. If there are 3 faulty items in the box, Find the probability that box is selected?
Solution:
𝑛 𝑆 =
12
4
𝐴: 𝐵𝑜𝑥 𝑖𝑠 𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑
𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
𝑛!
𝑟!
𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
𝑛!
𝑟! 𝑛 − 𝑟 !
𝑛 𝐴 =
3
0
9
4
+
3
1
9
3
= 126 + 252 = 378
𝑃 𝐴 =
𝑛 𝐴
𝑛 𝑆
=
378
495
= 0.76

*
The probability that two events 𝐴 & 𝐵 will occur in sequence.
𝑷 𝑨 ∩ 𝑩 = 𝑷 𝑨 × 𝑷 𝑩|𝑨
For Independent event the rule can be simplified to
𝑷 𝑨 ∩ 𝑩 = 𝑷 𝑨 × 𝑷 𝑩
Example 1:
The probability that a person A will be alive 15 years hence is
𝟓
𝟕
and other person B will
be alive 15 years hence is
𝟕
𝟗
. Find the probability that both will alive 15 years hence?
Solution:
Since P A =
5
7
P B =
7
9
Then the probability that both will alive 15 years is
𝑷 𝑨 ∩ 𝑩 = 𝑷 𝑨 × 𝑷 𝑩
𝑃 𝐴 ∩ 𝐵 =
5
7
×
7
9
=
5
9

*
Q: A die is rolled twice: Event 𝑬𝟏 is the appearance of even number of dots and event 𝑬𝟐 is
the appearance of more than 4 dots. Prove that:
𝑷 𝑬𝟏 ∩ 𝑬𝟐 = 𝑷 𝑬𝟏 × 𝑷 𝑬𝟐
Solution:
When dice is rolled then possible outcomes are
1, 2, 3, 4, 5, 6
𝑛 𝑆 = 6
Since 𝑬𝟏 is the event that the dots on the die are even then favorable outcomes are 2, 4, 6
𝑛 𝐸1 = 3
So, Probability
𝑃 𝐸1 =
𝑃 𝐸1
𝑛 𝑆
=
3
6
=
1
2
Now since 𝑬𝟐 is the event that the dot appear are more than four then favorable outcomes are
5 and 6
𝑛 𝐸2 = 2

*
*So, Probability
𝑃 𝐸2 =
𝑛 𝐸2
𝑛 𝑆
=
2
6
=
1
3
Since 𝐸1 𝑎𝑛𝑑 𝐸2 are not mutually exclusive
And the possible common outcome is 6
𝑛 𝐸1 ∩ 𝐸2 = 1
So, probability
𝑃 𝐸1 ∩ 𝐸2 =
𝑛 𝐸1 ∩ 𝐸2
𝑛 𝑆
=
1
6
Now,
𝑃 𝐸1 × 𝑃 𝐸2 =
1
2
×
1
3
=
1
6
Hence Proved,
𝑃 𝐸1 ∩ 𝐸2 = 𝑃 𝐸1 × 𝑃 𝐸2

• A discrete distribution describe the probability of
occurrence of each value of a discrete random variable.
• A discrete random variable is a random variable that has
countable values, such as a list of non-negative integers.
• With a discrete probability distribution, each possible
value of the discrete random variable can be associated
with a non-zero probability. Thus, a discrete probability
distribution is often presented in tabular form.
*

• With a discrete distribution, unlike with a continuous
distribution, you can calculate the probability that 𝑋 is exactly
equal to some value.
• For example, you can use the discrete Poisson distribution to
describe the number of costumer complaints within a day.
• Suppose the average number of complaints per day is 10 and
you want to know the probability of receiving 5, 10, and 15
customer complaints in a day.

𝒙 𝑷(𝑿 = 𝒙)
5 0.037833
10 0.125110
15 0.034718
*You can also view a
discrete distribution on a
distribution plot.

Resulting number Probability
1 1 out of 6
2 1 out of 6
3 1 out of 6
4 1 out of 6
5 1 out of 6
6 1 out of 6
Uniform
distribution

• Find the discrete probability for the number of heads on three
tosses of a fair coin, and the experiment is repeated 8 numbers
of time. Also draw a line.
*
Sr. no. Out comes of coins No. of heads
(X)
1 TTT 0
2 TTH 1
3 THT 1
4 THH 2
5 HTT 1
6 HTH 2
7 HHT 2
8 HHH 3

*
X f
𝑷 𝒙 =
𝒇
∑𝒇
0 1 1/8
1 3 3/8
2 3 3/8
3 1 1/8
∑𝑓 = 8 ∑𝑃 𝑥 = 1

*
Probability is used in such as
*Math
* Statistics
*Finance
* Gambling
* Science
*Machine and Artificial Intelligence

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