biometry

1. Probability
The term probability refers to indicate the
likelihood that some event will happen. For
example, „there is high probability that it will
rain tonight‟. We conclude probability based
on some observations or measurements.
m
Mathematically, probability is P n
Where „m‟ is the no of favourable trials and „n‟ is
the total no of trials.

2. PROBABILITY..
Sex (M:F)distribution in human being is 1:1. So
probability of a child to be born as M or F is
50% ie 0.5.

3. PROBABILIT
Introduction
A random variable is a variable whose actual
value is determined by chance operations.
When you flip a coin the outcome may be a
head or tail. Possibility of appearing head or
tail is 50%-50%. Probability of an albino
offspring from two heterozygous parents or
the probability of an individual being <6‟ tall
are also determined as outcomes of random
variables. Random variables may be discrete or
continuous.

4. DEFINITION
OF PROBABILITY
If a trial results in „n‟ exhaustive, mutually
exclusive and equally likely cases and „m‟ of
the favourable to an event A, then the
probability of p of the happening of A is given
by
Favourable number of cases
m
p
Total number of cases
n
This gives the numerical measure of probability.
Obviously “p” be a positive number not greater
than unity, so that 0≤ p ≤1.

5. PROBABILITY..
Assume an expt consist of one through of a 6
sided die. Possible results are 1, 2, 3, 4, 5 and
6. Each of those possible results is a simple
event. The probability of each of those events
is 1/6 ie P(E1) = P(E2) = P(E3) = P(E4) = P(E5) =
P(E6) =1/6.
This is shown in Table 1.

6. TABLE 1.
Observation
Event (Ei)
P(Ei)
1
E1
P(E1) = 1/6
2
E2
P(E2) = 1/6
3
E3
P(E3) = 1/6
4
E4
P(E4) = 1/6
5
E5
P(E5) = 1/6
6
E6
P(E6) =1/6

7. PROBABILITY DENSITY
Random
FUNCTION (PDF)1OR 2
variable:x
3
4
5
6
1/6
1/6
1/6
1/6
PROBABILITY
DISTRIBUTION
Density: f(x)
1/6
1/6
1.A fair 6-sided
dice is rolled with
the discrete
random variable
X representing
the number
obtained per roll.
Give the density
function for this
variable.
Values of X not listed in
the table are presumed
to be impossible to occur
and their corresponding
values of f are 0. For
example f(0.1)=0,
f(-3)=0. Notice that f 1

8. Density: f(x)
2
3
4
1
2
3
4
5
6
5
4
3
2
1
36
36
36
36
36
36
36
36
36
36
36
2.A fair 6-sided
dice is rolled twice
with the discrete
random variable X
representing the
sum of the
numbers obtained
on both rolls. Give
the pdf of this
variable.
5
6 7 8 9 10 11 12
PDF
Random
variable x
There are 6x6 different outcomes
possible,
so
each
has
a
probability of 1/36. The possible
values range from 2 to 12. While
2 (1+1) and 12 (6+6) can occur
only one way, all other values
occur at least two ways eg. A 3 is
either a 1 then a 2 or a 2 then a
1; a 4 is either a 1 then a 3, a 3
then a 1, or two 2‟s.

9. PROBABILITY DENSITY FUNCTION…
A sum of 10 in two
trial may appear in
three ways such as
5 + 5 = 10
ii) 4 + 6 = 10
iii) 6 + 4 = 10
i)
A sum of 7 in two trial
may appear in six
ways such as
i)
5+2=7
ii)
2+5=7
iii) 4 + 3 =7
iv)
3+4=7
v)
6+1=7
vi)
1+6=7

10. LAWS
OF
PROBABILITY
1. Additive
law of
probability
2. Multiplicative law of
probability

11. ADDITIVE LAW
OF
PROBABILITY
One can use either the density function table or graph to find the
probability of various outcomes. For example,
P(X = 10) = 3/36= 1/12 and
P(X = 10 or 11) = 3/36 + 2/36 = 5/36
10 and 11 are mutually exclusive events, so application of the general
addition law leads to summation of the individual probabilities.
This law is known as ADDITIVE LAW of probability

12. WHAT
IS MUTUALLY EXCLUSIVE
?????
Cases are said to be mutually exclusive if the
occurrence of one of them excludes the
occurrence of all the others.
For example, in tossing an unbiased coin, the
cases of appearing “head” and “tail” are
mutually exclusive.

13. MULTIPLICATIVE LAW
OF
PROBABILITY
Consider k sets of elements of size n1, n2,….nk.
If one element is randomly chosen from each
set, then the total no of different results is
n1n2n3…nk . Example: Consider 3 pens with
animals marked as
Pen 1: 1,2,3
Pen 2: A, B, C
Pen 3: x,y

14. MULTIPLICATIVE LAW
The
possible
triplets with
one animal
taken from
each pen
are
1Ax, 1Ay,
1Bx, 1By,
1Cx, 1Cy
OF
2Ax, 2Ay,
2Bx, 2By,
2Cx, 2Cy
PROBABILITY…
3Ax, 3Ay,
3Bx, 3By,
3Cx, 3Cy
The number
of possible
triplets is :
n!=
(3)(3)(2)=18

15. PERMUTATIONS
From a set of n elements, the number of ways
those n elements can be rearranged , ie put in
different orders, is the permutation of n
elements
Pn = n!
The symbol n! (factorial of n) denotes the
product of all natural numbers from 1 to n
n! = (1)(2)(3)….(n)
By definition 0! = 1.

16. PROBLEM 1:
In how many ways can three animals, x, y and z
be arranged in triplets ?
The number of permutations of n =3 elements
P(3) = 3! = (1)(2)(3) = 6. The six possible
triplets: xyz, xzy, yxz, yzx, zxy, zyx
More generally, we can define permutations of n
elements taken k at a time in particular order
as
Pn,k = n!
(n
k )!

17. EXAMPLE
In how many ways can three animals, x, y, z be
arranged in pairs such that the order in the
pairs is important (xz is different than zx)?
Pn,k = (3 3!2 )!= 6
The six possible pairs are: xy xz yx yz xz zy

18. MULTIPLICATIVE
Problem: A bag
contains
4
white and 5 red
balls. Two balls
are
drawn
successively at
random
from
the bag. What
is
the
probability that
both the balls
are white when
the
drawings
are made i)with
replacement?
ii)without
replacement?
LAW OF PROBABILITY..
Soln: Let A be the event that the
first ball is white and B be the event
that the second ball is also white.
i) Since the drawings are made with
replacement , the two events
become independent
Hence P(AB) = P(A) P(B) =
( 4/9)x(4/9) = 16/81
ii) Since the drawings are made
without replacement the events
become dependent.
Hence P(AB) = P(A) P(B/A) =
4/9x3/8 = 12/72

19. COMPOUND EVENTS
A compound event is an event composed of two or
more events. Consider two events A and B. The
compound event such that both events A and B
occur is intersection of the events, denoted by A ∩
B . The compound such that either event A or event
B occurs is called the union of events, denoted by A
U B. The probability of an intersection is P(A ∩ B)
and the probability of union is P(A U B).
Also P(A U B) = P(A) + P(B) – P(A∩ B)