Page 27 - DMTH404_STATISTICS

Unit 2: Methods of Enumeration of Probability

Notes
Example 4: We can extend the property P2 to the case of three events, i.e., we can show
that

P(A  B  C) = P(A) + P(B) + P(C) – P(A  B) – P(B  C ) – (C  A) + P(A  B  C) ... (5)
Denote B  C by H. Then A  B  C = A  H and by P2, P(A  B  C)
= P(A U H) = P(A) + P(H) – P(A  H). ... (6)
But P(H) = P(B  C) = P(B) + P(C) – P(B  C) ... (7)

and P(A  H) = P(A  (BC))
= P((A  B)(A  C))
= P(A  B) + P(A  C) – P{(A  B)  (A  C)}
= P(A  B) + P(A  C) – P(A  B  C) ... (8)

Substituting from (7) and (8) in (6) we get the required result. Also see Fig. 2.
Here are some simple exercises which you can solve by using P1-P7

2.2 Classical Definition of Probability

In the early stages, probability theory was mainly concerned with its applications to games of
chance. The sample space for these games consisted of a finite number of outcomes. These
simple situations led to a definition of probability which is now called the classical definiqon. It
has may limitations. For example, it cannot be applied to infinite sample space. However, it is
useful in understanding the concept of randomness so essential in the planning of experiments,
small and large-scale sample surveys, as well as in solving some interesting problems. We shall
motivate the classical definition with some examples. We shall then formulate the classical
definition and apply it to solve some simple problems.
Suppose we toss a coin. This experiment has only two possible outcomes : Head (H) and Tail (T).
If the coin is a balanced coin and is symmetric, there is no particular reason to expect that H is
more likely than Tor that T is more likely than H. In other words, we may assume that the two
outcomes H and T have the same probability or that they are equally likely. If they have the
same probability, and if the sum of the.two probabilities P(H) and P(T) is to be one, we inust
have P(H) = P(T) = 1/2.

Similarly, if we roll a symmetric, balanced die once, we should assign the same probability, viz.
116 to each of the six possible outcomes 1, 2, . . . ,6.
The same type of argument, when used for assigning probabilities to the results of drawing a
card from a well-shuffled pack of 52 playing cards leads us to say that the probability of drawing
any specified card is 1/52.

In general, we have the following :
Definition 2 : Suppose a sample space  has a finite number n of points  ,  , . . . ,  . The
1 2 n
classical definition assigns the probability l/n to each of these points, i.e.,
1
P{} = , j  1,....,n.
j n

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