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Probability
- Probability is a measure of how likely an event
is to occur.
- For example –
- Today there is a 60% chance of rain.
- The odds of winning the lottery are a million
to one.
0 <= p <= 1
0 = certain non-occurrence
1 = certain occurrence
- .5 = even odds
- .1 = 1 chance out of 10
Probability varies between 0 and 1
- If an event will NEVER happen, then the probability of the event is 0
or 0%., 0 indicates(0%) impossibility of event.
The higher the probability of an event, the more likely that the event will
occur.
For each outcome we will get probability
Sum of all the probability of sample space is equal to one, sample
space consists of outcomes.
- If an event is certain to happen, then the probability of the
event is 1 or 100%, 1 indicates(100%) certainty
- If an event is just as likely to happen as to not happen, then
the probability of the event is ½, 0.5 or 50%.
Impossible Unlikely Equal Chances Likely Certain
0 0.5 1
0% 50% 100%
½
- The probability of an event is written as:
P(event) = chance of event can occur
total number of outcomes
- An outcome(X) is a possible result of a
probability experiment
- When rolling a number cube, the possible
outcomes are 1, 2, 3, 4, 5, and 6
- When rolling a number cube, the event of rolling
an even number is 3 (you could roll a 2, 4 or 6).
What is the probability of getting heads when flipping a
coin?
P(heads) = 1 head on a coin = 1
total outcomes = 2 sides to a coin = 2
P(heads)= ½ = 0.5 = 50%
- What is the probability that the spinner will stop on
part A?
- What is the probability that the spinner will
stop on
(a) An even number?
(b) An odd number?
- What is the probability that the spinner
will stop in the area marked A?
B A
C D
3 1
2
A
C B
TRY THESE:
P(A)=1/4=25%
P(even)=1/
P(areaA)=1/
Probability Word Problem:
- Lawrence is the captain of his track team.
The team is deciding on a color and all
eight members wrote their choice down
on equal size cards. If Lawrence picks one
card at random, what is the probability
that he will pick blue?
Number of blues = 3
Total cards = 8
yellow
red
blue
blue
blue
green
black
black
3/8 or 0.375 or 37.5%
- Donald is rolling a number cube labeled
1 to 6. What is the probability of the
following?
a.) an odd number
odd numbers – 1, 3, 5
total numbers – 1, 2, 3, 4, 5, 6
b.) a number greater than 5
numbers greater – 6
total numbers – 1, 2, 3, 4, 5, 6
Let’s Work These Together
3/6 = ½ = 0.5 = 50%
1/6 = 0.166 = 16.6%
- What is the probability of spinning a number
greater than 1?
- What is the probability that a spinner with five
congruent sections numbered 1-5 will stop on an
even number?
- What is the probability of rolling a multiple of 2 with
one toss of a number cube?
TRY THESE:
1 2
3 4
P(>1)=3/4=75%
P(even)=2/5=40%
P(multiple of 2)=3/6=50%
Probability Properties
X is an outcome of an event
- A simple example is the tossing of a coin.
The two outcomes ("heads" and "tails") are both equally
probable; the probability of "heads" equals the probability of
"tails"; and since no other outcomes are possible, the
probability of either "heads" or "tails" is 1/2 (which could also
be written as 0.5 or 50%)
Ex: A Coin is tossed, find the probability head outcome.
P(X)= 1 / 2
Ex: A dice is thrown find the probability of each of the outcome
Probability(X)=Desired outcome/ total no. of outcomes
P= 1 / 6
Ex: tossing a coin twice find the probability of each outcomes.
possible outcomes "head-head", "head-tail", "tail-head", and "tail-tail"
outcomes.
Sample space S={HH,HT,TH,TT}
The probability of getting an outcome of “head-head" is 1 out of 4
outcomes or 1 / 4 or 0. 25 (or 25 %). Suppose X=heads
Total Probability=
X 2 1 0
P(x) ¼ 1 / 2 ¼
Probability of Event A
The probability of an event A is written as, P(A).
Probability of not Event A
The opposite or complement of an event A is the event
[not A] (that is, the event of A not occurring), often
denoted as
; its probability is given by
P(not A) = 1 − P(A).
Probability of Event A and not event A
Ex: Tossing a coin three times, write the sample space and
Find the probability of head outcome.
S={HHH, HHT,HTH,THH,HTT,TTH,THT,TTT}
Let X is random variable and X= count of heads in sample
space. When a coin tossed three times
First type of outcomes is X= 3 heads, then subsample (HHH)
Second type of outcomes X= 2 heads, then
Possibilities subsamples are (HHT,HTH,THH)
Third type of outcome X= 1 head,
then possibilities of subsamples (HTT,TTH,THT)
Forth type of outcome X= 0 heads, then possibilities (TTT)
Multiple outcomes ex: X= 2 (equal heads) then first use
intersection
=(( 1 / 2 x 1 / 2 x 1 / 2 ) +( 1 / 2 x 1 / 2 x 1 / 2 )+( 1 / 2 x 1 / 2 x 1 / 2 ))= 3 / 8
=( 1 / 2 x 1 / 2 x 1 / 2 )= 1 / 8 A Single outcome, use intersection i. e.
and then union
X=count of heads 3 2 1 0
P(x) 1 / 8 3 / 8 3 / 8 1 / 8
Probability distribution must be 1
Difference of random variables
X Difference between no. of heads and tail,
Ex: when two coins are tossed.
Probability distribution table for Difference, S{HH,HT,TH,TT}
H T Diff
X(HH,TT)=2 i.e. ((1/2x1/2)+(1/2x1/2))=1/
X(HT,TH)=0 i.e. ((1/2x1/2)+(1/2x1/2))=1/
Difference-X 2 0 sum
P(x) 1 / 2 1 / 2 1
Probability distribution table
Ex: If the outcome is head then 4 bits are sending per second, and if
the outcome is Tail then bits are not sent through the Tx line. When
three coins are tossed (i) find the bits sent for variable outcomes in the
sample space (ii) Find the probability distribution for the same.
Possibilities in
terms of no.(H)
Possibilities in
terms of no.(T)
send(+), not
sent(-) Difference
amount(X)
Possibilities
Bits sent/not
sent
HHH 3 0 12 Bits sent
HHT,THH,HTH 2 1 8 - 0 = 8 Bits sent
HTT,THT,TTH 1 2 4 - 0 = 4 Bits not sent
TTT 0 3 0 - 0 Bits not sent
P(HHH)=(1/2x1/2x1/2) = 1/
P(HHT,THH,HTH )=(1/2x1/2x1/2)+(1/2x1/2x1/2)+(1/2x1/2x1/2) = 3/
P(HTT,THT,TTH)=(1/2x1/2x1/2)+(1/2x1/2x1/2)+(1/2x1/2x1/2) = 3/
P(TTT)=(1/2x1/2x1/2) = 1/
Probability Distribution
X 12 8 4 0
P(x) 1 / 8 3 / 8 3 / 8 1 / 8
Total=1/8+3/8+3/8+1/8=
Mean of random variables
X x 1 x 2 x 3 x 4
P(x) P 1 P 2 P 3 P 4
For Discrete Random Variable: Expected value or
Mean of probability distribution or Mean of random variable,
Expected Value E(x)=X
1
P
1
+X
2
P
2
+X
3
P
3
+X
4
P
4
+--+X
n
P
n
mean ( ) E ( x ) xP ( x )
i
forall x
i
The Probability distribution of discrete random variable x, for Ex.
x 0 1 2
P(x) 0.16 0.48 0.
mean ( ) E ( x ) 0. 48 2 0. 36 1. 2
Ex: X is the real no. associated with sample space, When One dice is
thrown,
X 1 2 3 4 5 6
P(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6
X
2
1 4 9 16 25 36
E(X)=1/6+2/6+3/6+4/6+5/6+6/6=21/
Variance and Standard Deviation
Variance=E(x
2
)-(E(x))
2
Standard deviation( )= Varience
sample space S={1,2,3,4,5,6}
Varience 1. 71
E(x)=21/6=3.
E(x
2
)=1/6+4/6+9/6+16/6+25/6+36/6=91/6=15.
Variance=E(x
2
)-(E(x))
2
=2.
Standard deviation( )=
Ex: Two dice thrown at a time, if x outcome is denoted as the no. of
six, write Probability Distribution and find mean, variance, and
standard deviation
Dice1 dice2 X=No. of 6
6 6 not(1,2,3,4,5) 1
6not 6 1
6not 6not 0
P(6)=1/6, P(6not)=5/
P(6 6)=1/6x1/6=1/
P(6,6not)=(1/6x5/6)=1/
P(6not 6not)=(5/6x5/6)=25/
P(6 6not, 6not 6)=(1/6x5/6)+(5/6x1/6)=5/36+5/36=10/
X 0 1 2
P(x) 25 / 36 5 / 36 + 5 / 36 = 10 / 36 1 / 6 x 1 / 6 = 1 / 36
X
2
E(x)=(10/36)+(2x1/36)=12/36=1/
E(x
2
)=(10/36)+(4x(1/36))=14/
Variance(x)=E(x
2
)-(E(x))
2
Variance=(14/36-1/9)=(14-4)/36=10/36=0.
Varience 0. 527
Standard deviation( )=
Variance more, stability less in Discrete Random Variables
It is are of two types
- Cumulative Distribution Function
2. Probability Density (Distribution) Function
Probability Distributions
Continuous random variables: It is a range of random variable
Continuous random variable: Ex: Y= 2 , 2. 4 , 2. 8
- PDF is constant over the a,b interval
- CDF is continuously increasing we get the ramp function
CDF: Probability of previous outcome is added in to the
probability of next outcome is called Cumulative Density
Function (CDF), symbol Fx(x)
PDF: Probability is uniform for every outcome (ranndom
variable) is called Probability Density (in discrete it is
Probability Distribution) Function (PDF), Symbol f x(x)
3). Non decreasing
Properties
Cumulative distribution Function
probability of P( 2)=1/6+1/6=2/
probability of P( 1)=1/
probability of P( 3)=1/6+1/6+1/6=3/
Probability density (distribution) Function:
Properties:
(1)Finding a CDF where there is only one function in the pdf
(2)Finding a CDF where there is more than one function in the pdf
Conditional probability
Conditional probability is the probability of some event A , given that
another event B has occurred.
Conditional probability is written P(A/B), or P B
(A) and is read"the
probability of A , given B ".itis conditional probability A given B.
PB
PA B
P A B
If P ( A | B ) = P ( A ), then events A and B are said to be independent:
Also, in general, P ( A | B ) (the conditional probability of A given B) is not
equal to P ( B | A ).
Ex: Suppose that somebody secretly rolls two fair six-sided dice, and we
must predict the outcome (i.e sum of the two upward faces).
Example on conditional probability
- Moments of distribution is a set of parameters
- It consist of c and n values, depending on the values of c
and n, we can calculate various terms which are very
essential in statistics, and used in communication.
- C=0 then moments are w. r. t. 0 value, i.e. talking about
the raw or absolute moment.
th
moments;
- C= μ, then moments are w. r. t. population mean value,
it is about central moments.
st
moments;
nd
moment; and so on
Discrete random variable
2 2
EX xpx
C = 0 & n=1, 1
st
moment (absolute)
C = 0 & n=2, 2
nd
moment (absolute)
Moments and Expectation
Moments and Expectation: Continuous Random Variable
E X xf ( x ) dx
C 0 & n 1 , 1 moment ( absolute )
st
E X x f ( x ) dx
2 2
C 0 & n 22 moment ( absolute )
nd
Ex: Expectation E(X)
X= workout, the probability of workout in a week is given
as:
Find the expected value E(x)
or mean (μ) for the given
data
E(X) = μ = 2.
st
moment
nd
moment
2
E X
i i i
EX xpx
2 2
i i i
EX x px
( 1 p ). 0 p. 1 p
p p p
2 2
Ex: Variance(X)
p ( 1 p )
2
p p
2 2
Var ( X ) E ( X )( E ( X ))
Ex: Variance(X)
Check
Law of large numbers
X 1
, X 2
,------is an infinite sequence of i.i.d. of
The sample average converges to the expected value
n
n
n
EX EX E X
n
EX X X
E X
n n
n
2 1 2
n n
n
n
VarX VarX Var X
n
VarX X X
Var X
n n
n
2
2
2
2
2
2
1 2
( ) ( ) ( ) ( )
( )
2
2
2
for
n
Var X
P X
n
n
In Weak Law of Large Numbers states that:
Two different versions of the law of large numbers are described below; they are
called the strong law of large numbers , and the weak law of large numbers. Stated
for the case where X 1
, X 2
, ... is an infinite sequence of i.i.d. Lebesgue integrable
random variables with expected value E( X 1
) = E( X
2
) = ...= μ , both versions of the
law state that – with virtual certainty – the sample average
n
EX
( ) n
VarX
P ( X ) for 0. 1
n
