Transcript + P(B)

Introduction
Probability Theory was first used to
solve problems in gambling
 Blaise Pascal (1623-1662) - laid the
foundation for the Theory of
Probability
 Now this theory is used in business,
Science and industry
 Next we define-Experiment and Event

Simple Probability
An Experiment is an operation or a
process with a result or an outcome
which is determined by , or depends
on, Chance.
 Examples: (1) Tossing a coin.

(2) Tossing a die
 An Event is the outcome of an
experiment
 Example: Tossing a HEAD , getting a
SIX are events.

Definition of Probability

In an experiment resulting in n
equally likely outcomes, if m of these
outcomes are favour the occurrence
of an event E
Then the Probability of event E happening,
written as P(E), is defined as
P(E) =
No. of outcomes favourable to the occurance of E
=
Total number of equally likely outcomes
m
n
Example 1:
A two-digit number is written
down at random.
Find the probability that the
number will be
(i) smaller than 20
(ii) even
(iii) a multiple of 5
Example 1:
How many two digit numbers are there?
Is it 100 or 90 or 91
The correct answer is 90
Now we will find the probabilities.
How many numbers are less than 20 ? 10
How many are even numbers ? 45
How many are multiples of 5 ? 18
Probability
The possible outcomes is called the
Sample space (S)
Hence, the probability of an Event E,
P(E) = n(E)
n(S)
for any event E, 0  P(E)  1
If P(E) = 0, then the event cannot possibly occur
If P(E) = 1, then the event will certainly occur.
In the Probability Theory, an event is any subset of a
Sample Space
Sample Space
Possible outcomes of the following experiments
1. Tossing a Coin: S = [ H , T ]
2. Tossing a die : S = [ 1, 2, 3, 4, 5, 6 ]
3. Tossing two coins: S = [ HH, HT, TH, TT ]
4. Tossing two dice : S = [ (1,1),(1,2) ….. (6,5),(6,6)]
We can draw sample space for the above experiments
Example 2 :
Two dice are thrown together.
Find the probability that the sum of
the resulting numbers is
(a) odd
(b) even
(c) a prime number
(d) a multiple of 4
(e) at least 7
First we draw the sample space, then
using that we can find the probability .
Second die
First die
+
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
All the possible sums are displayed in the above diagram
This is called the sample space of this experiment
Example 2 :
We define the following:
Now, it is very easy to
A: the sum is odd
calculate the probabilities.
B: the sum is even
C: the sum is a prime number
D: the sum is a multiple of 4
E: the sum is at least 7
Total possible outcomes are 36. Hence n(S) = 36
n(A) = 18
A and B are complementary events. n(B ) = 36 - n(A)
n(C) = 15
n(D) = 9
n(E) = 21
Answers:
18
1
( a ) P(A) =

36
2
(b) P(B) =
( c) P(C) =
( d ) P(D) =
( e)
P(E) =
1
1
1 - P(A) = 1 
2
2
15
5

36 12
9
1

36
4
21
7

36 12
Hence P(E’) = 1 - P(E) , where E’ is the
complement of E
Example 3 :
Box A contains 4 pieces of paper numbered 1,2,3,4
Box B contains 2 pieces of paper numbered 1,2.
One piece of paper is removed at random from each box
The sample space is as follows
2
1
1
2
3
Box A
4
Another way to illustrate the possible outcome
TREE DIAGRAM
Box B
1
Box A
2
3
1
(2,1)
2
1
(2,2)
2
1
4
2
(1,1)
(1,2)
2
1
Outcome
(3,1)
(3,2)
(4,1)
(4,2)
Hence n (S) = 8
A coin is tossed three times.
Display all the outcomes using a tree diagram
find the probability of getting
(i) three heads (ii) exactly two heads (iii) at least two heads
H
[H,H,H]
H
H
T
H
T
n (S) = 8
T
H
T
[H,H,T]
[H,T,H]
[H,T,T]
H
[T,H,H]
T
[T,H,T]
H
[T,T,H]
T
T
Now it is very easy to find the probabilities.
[T,T,T]
Adding Probabilities - Mutually Exclusive Events
Two events A and B are said to be Mutually Exclusive(ME)
if the occurance of one event will not affect the occurance
of the other event.
Set theoritically
A B  
Hence these two events A and B cannot occur
simultaneously.
If you want to calculate the probability of A or B, then
P(A or B )= P(AUB) = P(A) + P(B)
Example: A - getting an odd number
B - getting an even number, while tossing a die once
Also, P(AUBUC) = P(A) + P(B) + P(C)
Example 4 :
The probabilities of three teams L, M
and N, winning a football competition
are 1/4 , 1/8 and 1/10 respectively.
Calculate the probability that
(i) either L or M wins,
(ii) neither L nor N wins.
Example 4 :

We assume that only one team can win,
so the events are mutually exclusive.
(i) P( L or M wins) = P(Lwins) + P(M wins)
= 1/4 + 1/8 = 3/8
(ii) P(L or N wins) = 1/4 + 1/10 = 7/20
P(neither L nor N wins ) = 1 - P(L or N wins)
= 1 - 7/20 = 13/20
Note:
“Branches” of a “Probability Tree” represent outcomes
which are mutually exclusive
Example 5
Consider the experiment,
Tossing a die once
let A - getting an odd number [ 1,3,5]
B - getting a prime number [2,3,5]
Here, A intersection B is not empty
P(A) = 3/6 = 1/2, P(B) = 3/6 = 1/2
A  B = [3,5] , P(A  B) = 2/6 = 1/3
P(AUB) = P(A) + P(B) - P(A  B)
= 3/6 + 3/6 - 2/6 = 4/6 = 2/3
TREE DIAGRAM
Box A
Box B
1
(1,1)
2
(1,2)
1
(2,1)
2
1
(2,2)
(3,1)
2
(3,2)
1
2
3
4
Outcomes
1
(4,1)
2
(4,2)
Hence n (S) = 8
Now, we go back to the same
example
Box A contains 4 pieces of paper numbered
1,2,3,4
Box B contains 2 pieces of paper numbered
1,2.
One piece of paper is removed at random
from each box
If we replace the numbers 1,2,3..
by the corresponding probabilities, we get
TREE DIAGRAM
Box B ( 1 )
1
( )
4
Box A
1
( )
4
1
1
2
1
2
3
4
2
1
( )
4
2
1
1
( )
4
2
1
2
1
( )
2
1
( )
2 1
1
( )
2
1
( )
2
Probability
P(1,1)= 1/8
P(1,2)=1/8
P(2,1)=1/8
( )
2
1
( )
2
P(2,2)=1/8
1
( )
2
P(3,2)=1/8
P(3,1)=1/8
P(4,1)=1/8
P(4,2)=1/8
To find P(1,2), we multiply along the branches
Multiplying Probabilities - Probability Tree
The probability that two events, A
and B, will both occur, written as
P(A occurs and B occurs) or
simply P(A and B), is given by
P(A and B ) = P(A) x P(B)
Note: we have to multiply the probabilities along the
branch,
Consider the following example:
Suppose in a bag, there are 5 blue and 3 yellow marbles.
A marble is drawn at random from the bag, the colour
in noted and the marble is replaced. A second marble
2 st draw
is then drawn.
5
8
1 st draw
5
8
B
3
8
5
8
Y
B
3
8
3
8
This shows the
Probability Tree
Y
B
Y
Since the first marble drawn is replaced, the total number
of marbles in the bag remains the same for the second draw.
Hence, we can say that the results of the two draws
are independent and the two outcomes from each of
the two draws are independent events.
If the marble is not replaced, then the probability
of selecting a marble from the second draw is
affected. This kind of events are called
B
dependent events
5
Probability Tree
diagram for
dependent events
5
8
7
B
5
7
3
8
Y
3
7 Y
B
3
7
Y
A garden has three flower beds. The first bed has
20 daffodils and 20 tulips, the second has 30
daffodils and 10 tulips and the third has 10
daffodils and 20 tulips.
A flower bed is to be chosen by throwing a die
which has its six faces numbered 1,1,1,2,2,3. If
the die shows a
‘1’, the first flower bed is chosen, if it shows a
‘2’ the second bed is chosen and so on.
A flower is then to be picked at random from the
chosen bed.
Copy and complete the Probability Tree:
Bed 1
1
( )
2
( 1)
3
(
1
6
)
( 1)
2
daffodil
(1 )
2
Tulip
[20]
[40]
daffodil
3
( )
Bed 2
4
Tulip
( 1)
4
daffodil
1
( )
Bed 3
3
Tulip
( 2)
3
[20]
[30]
[40]
[10]
[10]
[30]
[20]
Sample Space S = [ 1,1,1,2,2,3]
1 1 1 3 1 1 5
x  x  x 
Prob. of picking a daffodil =
2 2 3 4 6 3 9
SUMMARY
 In an experiment which results in n equally
likely outcomes, if m of these outcomes
favour the occurrence of an event E,
then probability of the event E is given by
P(E) = m / n, also 0  P(E)  1
 The Sample Space refers to the set of all
the possible outcomes of an experiment.
 An event is any subset of the sample space
SUMMARY
If A and B are Mutually Exclusive,
then P(A or B) = P(A) + P(B)
If A and B are independent,
then P(A and B) = P(A) x P(B)