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venn diagram and utility to probability

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1) a) see the entire shaded region in the following Venn diagram is A while the mustard colour
portion is
AB
and the pink colour is
AB
Summing the two disjoint regions, we get A. The left over circular region or unshaded circular
region is
AB
(b) Adding the remaining circular portion in the above diagram
AB
, it will be nothing but
AB
.
So, the Venn diagram of
AB
is
(c) See that
,&A B A B A B

are disjoint sets in which
A B A B A
... (*) and
A B A B B
More clearly, the mustard and pink portions union is A while the pink and violet colour
portions union is B in the above diagram.
The entire shaded region is
AB
.
That is,
A B A B A B A B

...... (**)
In view of the equation (*), we can write this as
Or,
A A B A B
.
The suitable Venn diagram is
(2)(a) Observe that
,A B A B
are disjoint sets such that their union is A.
So, we can write
P A P A B A B
Since these sets have empty intersection, the union can be replaced by addition in the case of
numbers.
That is,
P A P A B P A B

From this, we can confirm that
P A P A B
.

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The equality holds when
0P A B
(b) We have seen in the result 1(c) that,
A A B A B
and
,A A B
are disjoint sets.
So, as in the result (a) here, we can write
P A B P A A B
In view of the disjoint sets
,A A B
, the union of the sets can be replaced by addition when
we are dealing with real numbers.
That is,
P A B P A P A B

...... (***)
More precisely,
P A P A B
(3) Consider the equation (*) in the result (1)(c).
That is,
P A P A B P A B

From this, we can write
P A B P A P A B

(4) Suppose S is the sample space of an experiment. All the events are seen in the rectangle of the
Venn diagram. If all the events in the rectangle put together represented by E, then
1PE
where E
refers all the events in the sample space.
If A and B are particular events in the sample space as shown in the solutions (1)(a),(b), (c), then, we
can see that the unshaded region in (1) (c) is
AB

.
That is, the green shaded region is the complement of
AB
denoted by
AB

.
So,
1P E P A B P A B

...... (****)
But, we have
A A B A B
and
B A B A B
While all these sets are disjoint, we have
P A P A B P A B

and
P B P A B P A B

P A P B P A B P A B P A B P A B

Using (**) of result (1c), we can write
P A P B P A B P A B
In other words,
P A B P A P B P A B
Using this in (****), we get
1P A B P A P B P A B

which is the required result.
5) We have seen that
P A P B P A B P A B P A B P A B

2P A P B P A B P A B P A B

6a) we have seen that
A A B A B
such that
AB
and
AB
are disjoint sets.
So,
P A P A B P A B

So,
P A B P A
Similarly,
B A B A B
gives
P A B P B
Adding the respective sides of these inequalities, we get
2P A B P A P B
From this, we can write
P A B P A P B
b) Suppose A and B are two events in the sample space.

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Then, we have
1P A B P A B


1P A P B P A B P A B

From this, we can write
1P A P B P A B
Applying the negative sign throughout, we get
1P A P B P A B
Taking the 1
st
two terms to the other side, we are left with
1P A B P A P B
7) a) we have seen that
A B A B A B A B

and all the three sets are disjoint.
Also, we have
A A B A B
So, the above equation becomes
A A B A B
b) If E is the sample space or the universal set, then, we can write
B B E
B A A
B A B A
A B A B
8) suppose
12
, ,...,
n
E E E
are the sample events in the sample space E.
Then, we have
1 2 1 2 1 2
P E E P E P E P E E
From this, we can say that
1 2 1 2
P E E P E P E
while
12
0P E E
...... (Y)
In view of mathematical induction, suppose
1
1
k
k
ii
i
i
P E P E



...... (M)
Also, suppose
1
k
it
i
EE
...... (X)
When n = k+1, we can write
1
1
11
kk
i i k
ii
P E P E E

Using (X) in this, we can say that
1
1
1
k
i t k
i
P E P E E



Using (Y) in this, we can write
11t k t k
P E E P E P E


Again, substituting (X), we can say that
1
11
11
kk
i i k t k
ii
P E P E E P E P E


So, (M) allows us to write
1 2 1 1
1
...
k
k k i k
i
P E E E E P E P E


Or,
1 2 1 1 2 1
... ...
k k k k
P E E E E P E P E P E P E

Hence the result.
9) Suppose the probability of the occurrence of an event A is a = P (A).
Then the non occurrence of the event has the probability of A is given by
1P A P A b
In a more explanatory way, we follow that
1P A P A

while these two are disjoint sets or
exclusive events.
Or,
1ab
This can now be used as
1
aa
P A a
ab
10) a) only F can be written as
E F G

b) E and F but not G can be written as
E G G
c) At least one event occurs
E F G

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Anonymous
I was struggling with this subject, and this helped me a ton!

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