Page 25 - DMTH404_STATISTICS

Unit 2: Methods of Enumeration of Probability

Now here is an important remark. Notes

Remark 1 : If A = , P() = 0. However, P(A) = 0 does not, in general, imply that A is the empty set.
j
For example, consider the assignment (i) of E1. You must have already shown that it is valid. If
A = { ,  ) . P(A) = 0 but A is not empty.
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Similarly P() = 1. But if P(B) = 1, does it follow that B =  ? No. Can you think of a Probability
on a Discrete sample counter example ? What about El) i) again ? If we take B = {  ,  ,  ,  ,  },
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then P(B) = 1 but B  . In this connection, recall that the empty set and the whole space were
called the impossible event and the sure event respectively. In future, an event A with probability
P(A) = 0 will be called a null event and an event B of probability one, will be called an almost
sure event.
This remark brings out the fact that the impossible event is a null event but that a null event is
not the impossible event. Similarly, the sure event is an almost sure event but an almost sure
event is not necessarily the sure event.
Let us take up another property now.
P2 : P(AB) = P(A) + P(B) – P(AB).

Proof : Recall that according to the definition, P(AB) is the sum of the probabilities attached
to the points of AB, each point being considered only once. However, when we compute
P(A) + P(B), a point in A  B is included once in the computation of P(A) and once in the
computation of P(B). Thus, the probabilities of points in A n B get added twice in the computation
of P(A) + P(B). If we sdbtract the probabilities of all points in A  B, from P(A) + P(B), then we
shall be left with P(A  B), i.e.,

P(AB) = P(A) + P(B) –  P{ }

j
j   A  B
The last term in the above relation is, by definition, P(AB). Hence we have proved P2. We now
list some properties which follow from P1 and P2.
P3 : If A and B are disjoint events, then

P(A B) = P(A) + P(B)
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P4 : P(A ) = 1 – P(A)
P5 : P(AB)  P(A) + P(B)
Why don’t you try to prove these yourself? That’s what we suggest in the following exercise.
We continue with the list of properties.

P6 : If A  B, then P(A)  P(B).
Proof: If A  B, A and B – A are disjoint events and their union, A  (B – A) is B. Also see Fig. 1.
Hence by P3, .

P(B) = P(A(B – A)) = P(A) + P(B – A)
Since by P1, P(B – A)  0, P6 follows from the above equation.
Now let us take a look at P5 again.

The inequality P(A  B)  P(A) + P(B) in P5 is sometimes called Boole’s inequality. We claim that
equality holds in Boole’s inequality if A  B is a null event. Do you agree?
An easy induction argument leads to the following generalisation of P5.

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