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THE RELATIVE FREQUENCY DEFINITION OF PROBABILITY:INDEPENDENT EVENTS

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MTH001 ­ Elementary Mathematics
LECTURE # 32:
·  Independent and Dependent Events
·  Multiplication Theorem of Probability for Independent Events
·  Marginal Probability
Before we proceed the concept of independent versus dependent events, let us review the
Addition and Multiplication Theorems of Probability that were discussed in the last lecture.
To this end, let us consider an interesting example that illustrates the application of both of
these theorems in one problem:
EXAMPLE:
A bag contains 10 white and 3 black balls. Another bag contains 3 white and 5 black balls.
Two balls are transferred from first bag and placed in the second, and then one ball is taken
from the latter.
What is the probability that it is a white ball?
In the beginning of the experiment, we have:
Colour of
No. of
No. of
Ball
Balls in Bag A
Balls in Bag B
White
10
3
Black
3
5
Total
13
8
Let A represent the event that 2 balls are drawn from the first bag and transferred to the
second bag. Then A can occur in the following three mutually exclusive ways:
A1 = 2 white balls are transferred to the second bag.
A2 = 1 white ball and 1 black ball are transferred to the second bag. 13
⎜  ⎟.
2
A3 = 2 black balls are transferred to the second bag.
⎝  ⎠
Then, the total number of ways in which 2 balls can be drawn out of a total of 13 balls is 10
⎜  ⎟.
2
⎝  ⎠
And, the total number of ways in which 2 white balls can be drawn out of 10 white balls is
Thus, the probability that two white balls are selected from the first bag containing 13 balls
(in order to transfer to the second bag) is
10 ⎞ ⎛13 45
P(  A1 ) = ⎜  ⎟ ÷ ⎜  ⎟ =
2 ⎟ ⎜ 2  78 ,
⎝  ⎠ ⎝  ⎠
Similarly, the probability that one white ball and one black ball are selected from the first bag
containing 13 balls (in order to transfer to the second bag) is
10 ⎞ ⎛ 3⎞ ⎛13 30
P(  A2 ) = ⎜  ⎟ ⎜ ⎟ ÷ ⎜  ⎟ =  ,
1 ⎟ ⎜1 ⎟ ⎜ 2  78
⎝  ⎠⎝ ⎠ ⎝  ⎠
And, the probability that two black balls are selected from the first bag containing 13 balls (in
order to transfer to the second bag) is
3 ⎞ ⎛13 3
P(  A3 ) = ⎜ ⎟ ÷ ⎜  ⎟ =  .
2 ⎟ ⎜ 2  78
⎝ ⎠ ⎝  ⎠
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MTH001 ­ Elementary Mathematics
AFTER having transferred 2 balls from the first bag, the second bag contains
i) 5 white and 5 black balls (if 2 white balls are transferred)
Colour of
No. of
No. of
Ball
Balls in Bag A
Balls in Bag B
White
10 ­ 2 = 8
3+2=5
Black
3
5
Total
13 ­ 2 = 11
8 + 2 = 10
Hence: P(W/A1) = 5/10
ii)
4 white and 6 black balls
(if 1 white and 1 black ball are transferred)
Colour of
No. of
No. of
Ball
Balls in Bag A
Balls in Bag B
White
10 ­ 1 = 7
3+1=4
Black
3­1=2
5+1=4
Total
13 ­ 2 = 11
8 + 2 = 10
Hence: P(W/A2) = 4/10
iii)
3 white and 7 black balls
(if 2 black balls are transferred)
Colour of
No. of
No. of
Ball
Balls in Bag A
Balls in Bag B
White
10
3
Black
3­2=1
5+2=7
Total
13 ­ 2 = 11
8 + 2 = 10
Hence: P(W/A3) = 3/10
Let W represent the event that the WHITE ball is drawn from the second bag after
having transferred 2 balls from the first bag.
Then P(W) = P(A1W) + P(A2W) + P(A3W)
Now  P(A1 W) = P(A1)P(W/A1)
= 45/78 × 5/10
= 15/52
P(A2 W) = P(A2)P(W/A2)
= 30/78 × 4/10
= 2/13,
and
P(A3 W) = P(A3)P(W/A3)
= 3/78 × 3/10
= 3/260.
Hence the required probability is
P(W)
= P(A1W) + P(A2W) + P(A3W)
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MTH001 ­ Elementary Mathematics
= 15/52 + 2/13 + 3/260
= 59/130
= 0.45
Next, we discuss the concept of INDEPENDENT EVENTS:
INDEPENDENT EVENTS:
Two events A and B in the same sample space S, are defined to be independent (or
statistically independent) if the probability that one event occurs, is not affected by whether
the other event has or has not occurred, that is
P(A/B) = P(A) and P(B/A) = P(B).
It then follows that two events A and B are independent if and only if
P(A B) = P(A) P(B)
and this is known as the special case of the Multiplication Theorem of Probability.
RATIONALE:
According to the multiplication theorem of probability, we have:
P(A B) = P(A) . P(B/A)
Putting P(B/A) = P(B), we obtain
P(A B) = P(A) P(B)
The events A and B are defined to be DEPENDENT if P(AB) P(A) × P(B).
This means that the occurrence of one of the events in some way affects the probability of
the occurrence of the other event. Speaking of independent events, it is to be emphasized
that two events that are independent, can NEVER be mutually exclusive.
EXAMPLE:
Two fair dice, one red and one green, are thrown.
Let A denote the event that the red die shows an even number and let B denote the event
that the green die shows a 5 or a 6. Show that the events A and B are independent.
The sample space S is represented by the following 36 outcomes:
S = {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6);
(2, 1), (2, 2), (2, 3), (2, 5), (2, 6);
(3, 1), (3, 2), (3, 3), (3, 5), (3, 6);
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6);
(5, 1), (5, 2), (5, 3), (5, 5), (5, 6);
(6, 1), (6, 2), (6, 3), (6, 5), (6, 6) }
Since
A represents the event that red die shows an even number, and B represents the event that
green die shows a 5 or a 6,
Therefore A B represents the event that red die shows an even number and green die
shows a 5 or a 6.
Since A represents the event that red die shows an even number, hence P(A) = 3/6.
Similarly, since B represents the event that green die shows a 5 or a 6, hence P(B) = 2/6.
Now, in order to compute the probability of the joint event A B, the first point to
note is that, in all, there are 36 possible outcomes when we throw the two dice together, i.e.
S = {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6);
(2, 1), (2, 2), (2, 3), (2, 5), (2, 6);
(3, 1), (3, 2), (3, 3), (3, 5), (3, 6);
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6);
(5, 1), (5, 2), (5, 3), (5, 5), (5, 6);
(6, 1), (6, 2), (6, 3), (6, 5), (6, 6) }
The joint event A B contains only 6 outcomes out of the 36 possible outcomes.
These are (2, 5), (4, 5), (6, 5), (2, 6), (4, 6), and (6, 6).
P(A B) = 6/36.
and
Now
P(A) P(B)
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MTH001 ­ Elementary Mathematics
= 3/6 × 2/6
= 6/36
= P(A B).
Therefore the events A and B are independent.
Let us now go back to the example pertaining to live births and stillbirths that we
considered in the last lecture, and try to determine whether or not sex of the baby and
nature of birth are independent.
EXAMPLE :
Table-1 below shows the numbers of births in England and Wales in 1956 classified
by (a) sex and (b) whether live born or stillborn.
Table-1
Number of births in England and Wales in 1956 by sex and whether live- or still born.
(Source Annual Statistical Review)
Liveborn
Stillborn
Total
Male
359,881 (A)
8,609 (B)
368,490
Female  340,454 (B)
7,796 (D)
348,250
Total
700,335
16,405
716,740
There are four possible events in this double classification:
·  Male live birth,
·  Male stillbirth,
·  Female live birth, and
·  Female stillbirth.
The corresponding relative frequencies are given in Table-2.
Table-2
Proportion of births in England and Wales in 1956 by sex and whether live- or stillborn.
(Source Annual Statistical Review)
Liveborn
Stillborn
Total
Male
.5021
.0120
.5141
Female
.4750
.0109
.4859
Total
.9771
.0229
1.0000
As discussed in the last lecture, the total number of births is large enough for these relative
frequencies to be treated for all practical purposes as PROBABILITIES.
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MTH001 ­ Elementary Mathematics
The compound events `Male birth' and `Stillbirth' may be represented by the letters M
and S.
If M represents a male birth and S a stillbirth, we find that
n(M and S )
8609
=
= 0.0234
n (M )
368490
This figure is the proportion ­­ and, since the sample size is large, it can be regarded as the
probability ­­ of males who are still born ­­ in other words, the CONDITIONAL probability of
a stillbirth given that it is a male birth. In other words, the probability of stillbirths in males.
The corresponding proportion of stillbirths among females is
7796
= 0.0224.
348258
These figures should be contrasted with the OVERALL, or UNCONDITIONAL, proportion of
stillbirths, which is
16405
= 0.0229.
716740
We observe that the conditional probability of stillbirths among boys is slightly HIGHER than
the overall proportion. Where as the conditional proportion of stillbirths among girls is slightly
LOWER than the overall proportion. It can be concluded that sex and stillbirth are
statistically DEPENDENT, that is to say, the SEX of a baby yet to be born has an effect,
(although a small effect), on its chance of being stillborn. The example that we just
considered point out the concept of MARGINAL PROBABILITY.
Let us have another look at the data regarding the live births and stillbirths in
England and Wales:
Table-2Proportion of births in England and Wales in 1956 by sex and whether live- or
stillborn. (Source Annual Statistical Review)
Liveborn
Stillborn
Total
Male
.5021
.0120
.5141
Female
.4750
.0109
.4859
Total
.9771
.0229
1.0000
And, the figures in Table-2 indicate that the probability of male birth is 0.5141, whereas the
probability of female birth is 0.4859.Also, the probability of live birth is 0.9771, where as the
probability of stillbirth is 0.0229.
And since these probabilities appear in the margins of the Table, they are known as
Marginal Probabilities. According to the above table, the probability that a new born baby is
a male and is live born is 0.5021 whereas the probability that a new born baby is a male and
is stillborn is 0.0120.Also, as stated earlier, the probability that a new born baby is a male is
0.5141, and, CLEARLY, 0.5141 = 0.5021 + 0.0120.
Hence, it is clear that the joint probabilities occurring in any row of the table ADD UP to yield
the corresponding marginal probability.
If we reflect upon this situation carefully, we will realize that this equation is totally in
accordance with the Addition Theorem of Probability for mutually exclusive events.
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MTH001 ­ Elementary Mathematics
P(male birth)
= P(male live-born or male stillborn)
= P(male live-born) + P(male stillborn)
= 0.5021 + 0.0120
= 0.5141
Another important point to be noted is that:
Conditional Probability
Joint Probability
Marginal Probability
EXAMPLE:
P(stillbirth/male birth)
P(male birth and stillbirth)/P(male birth)
=0.0120/0.5141
= 0.0233
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Table of Contents:
  1. Recommended Books:Set of Integers, SYMBOLIC REPRESENTATION
  2. Truth Tables for:DE MORGAN’S LAWS, TAUTOLOGY
  3. APPLYING LAWS OF LOGIC:TRANSLATING ENGLISH SENTENCES TO SYMBOLS
  4. BICONDITIONAL:LOGICAL EQUIVALENCE INVOLVING BICONDITIONAL
  5. BICONDITIONAL:ARGUMENT, VALID AND INVALID ARGUMENT
  6. BICONDITIONAL:TABULAR FORM, SUBSET, EQUAL SETS
  7. BICONDITIONAL:UNION, VENN DIAGRAM FOR UNION
  8. ORDERED PAIR:BINARY RELATION, BINARY RELATION
  9. REFLEXIVE RELATION:SYMMETRIC RELATION, TRANSITIVE RELATION
  10. REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION
  11. RELATIONS AND FUNCTIONS:FUNCTIONS AND NONFUNCTIONS
  12. INJECTIVE FUNCTION or ONE-TO-ONE FUNCTION:FUNCTION NOT ONTO
  13. SEQUENCE:ARITHMETIC SEQUENCE, GEOMETRIC SEQUENCE:
  14. SERIES:SUMMATION NOTATION, COMPUTING SUMMATIONS:
  15. Applications of Basic Mathematics Part 1:BASIC ARITHMETIC OPERATIONS
  16. Applications of Basic Mathematics Part 4:PERCENTAGE CHANGE
  17. Applications of Basic Mathematics Part 5:DECREASE IN RATE
  18. Applications of Basic Mathematics:NOTATIONS, ACCUMULATED VALUE
  19. Matrix and its dimension Types of matrix:TYPICAL APPLICATIONS
  20. MATRICES:Matrix Representation, ADDITION AND SUBTRACTION OF MATRICES
  21. RATIO AND PROPORTION MERCHANDISING:Punch recipe, PROPORTION
  22. WHAT IS STATISTICS?:CHARACTERISTICS OF THE SCIENCE OF STATISTICS
  23. WHAT IS STATISTICS?:COMPONENT BAR CHAR, MULTIPLE BAR CHART
  24. WHAT IS STATISTICS?:DESIRABLE PROPERTIES OF THE MODE, THE ARITHMETIC MEAN
  25. Median in Case of a Frequency Distribution of a Continuous Variable
  26. GEOMETRIC MEAN:HARMONIC MEAN, MID-QUARTILE RANGE
  27. GEOMETRIC MEAN:Number of Pupils, QUARTILE DEVIATION:
  28. GEOMETRIC MEAN:MEAN DEVIATION FOR GROUPED DATA
  29. COUNTING RULES:RULE OF PERMUTATION, RULE OF COMBINATION
  30. Definitions of Probability:MUTUALLY EXCLUSIVE EVENTS, Venn Diagram
  31. THE RELATIVE FREQUENCY DEFINITION OF PROBABILITY:ADDITION LAW
  32. THE RELATIVE FREQUENCY DEFINITION OF PROBABILITY:INDEPENDENT EVENTS