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STA301 – Statistics and Probability
LECTURE NO 21:
•
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
Ball
White
No. of
Balls in Bag A
10
No. of
Balls in Bag B
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
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 
 2
10 
And, the total number of ways in which 2 white balls can be drawn out of 10 white balls is 
 2  .
 
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
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
Ball
White
No. of
Balls in Bag A
10 – 2 = 8
No. of
Balls in Bag B
3+2=5
Black
3
5
Total
13 – 2 = 11
8 + 2 = 10
Hence: P(W/A1) = 5/10
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STA301 – Statistics and Probability
ii)
4 white and 6 black balls
(if 1 white and 1 black ball are transferred)
Colour of
Ball
White
No. of
Balls in Bag A
10 – 1 = 7
No. of
Balls in Bag B
3+1=4
Black
3–1=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
Ball
White
No. of
Balls in Bag A
10
No. of
Balls in Bag B
3
Black
3–2=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(A1W) + P(A2W) + P(A3W)
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(A1W) + P(A2W) + P(A3W)
= 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(AB)  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.
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STA301 – Statistics and Probability
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).
and
P(A  B) = 6/36.
Now
P(A) P(B)
= 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
Male
359,881 (A)
Female 340,454 (B)
Total
700,335
Stillborn
Total
8,609 (B)
7,796 (D)
16,405
368,490
348,250
716,740
There are four possible events in this double classification:
•
Male live birth,
•
Male stillbirth,
•
Female live birth, and
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STA301 – Statistics and Probability
•
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
.9771
.0229
1.0000
Total
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.
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
.9771
.0229
1.0000
Total
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
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STA301 – Statistics and Probability
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.
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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