Page 50 - DMTH404_STATISTICS

Statistics

Notes We are sure that you will be able to do this simple exercise.

Task Find the sample spaces of the following experiments
(a) Rolling two dice
(b) Drawing two cards from a pack of 52 playing cards, with replacement.

Do you remember the definition of the Cartesian product of n(n  3) sets? We say that the
Cartesian product
S × S × ... × S = {(x ...., x ) | x  S, j = l ,... n}.
1 2 n 1 n j j
Now, if S , S , . . . . S represent the sample spaces corresponding to repetitions  ,  , . . . ,  of
1 2 n 1 2 n
the same experiment E, then the Cartesian product S × S × . . . × S represents the sample space
1 2 n
for n repetitions or n trials of the experiment .
[(H, T), (T, H), (T, T)] which is the Cartesian product of {H, T} with itself. Suppose the Probability
on a Discrete Sample coin is unbiased so that P{H} = P{T} = ID for both the first and the second
toss. Since the coin is unbiased, we may regard the four points in  as equally likely and assign
probability 1/4 to each one of them. However, another way of looking at this assignment is to
assume that the results in the two tosses are independent. Mor specifically, we may consider
specifying P{(H, H)} say, by the multiplication rul Jav ailable to us under independence, i.e., we
may take

1 1 1
P{H, H} = P{H} . P{H} = .  .
2 2 4
and make similar calculations for other pomts.

When such a situation holds, we say that the two tosses or the two trials of tossing the coin are
independent. This is equivalent to saying that the events Head on first toss and Head on second
toss are independent and that we may make similar statements about the other points also. The
following example illustrates the method of defining probabilities on the product spaces when
we are unable (or unwilling) to assume equally likely outcomes.

Example 26: Suppqe this successive units manufactured by a machine are such that each
unit has probability p of being defective (D) and (1 – p) of being good (G). We examine three
units manufactured by this machine. The sample space for this experiment is the Cartesian
product S × S × S , where S = S = S = {D, G}, i.e.,
1 2 3 1 2 3
 = [(D, D, D), (D, D, G), (D, G, D), (G, D, D), (G, G, D), (G, D, G), (D, G, G), (G, G, G)].
The statement that “the successive units are independent of each other’’ is interpreted by assigning
probabilities to points of  by the product rule. In particular,
P{(D, D, D)} = P(D) P(D) P(D) = p ,
3
P{(D, D, G)} = P(D) P(D) P(G) = p2q
= P{(D, G, D)} = P{(G, D, D)},

2
P{(G, G, D)} = P(G) P(G) P(D) = (1 – p) p
and lastly,
3
P((G, G, G)) = P(G) P(G) P(G) = (1 – p) .

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