Page 36 - Probability and Statistical Inference

1. Nations of Probability 13

question (iii), we obtain P(B | A) = P(A ∩ B)/P(A) = 1/3, that is there is one-
in-three chance that the randomly selected household subscribes for the maga-
zine M2 given that this household already receives the magazine M1. !

In the Example 1.4.3, P(A ∩ B) was given to us, and so we
could use (1.4.3) to find the conditional probability P(B | A).

Example 1.4.4 Suppose that we have an urn at our disposal which contains
eight green and twelve blue marbles, all of equal size and weight. The follow-
ing random experiment is now performed. The marbles inside the urn are mixed
and then one marble is picked from the urn at random, but we do not observe its
color. This first drawn marble is not returned to the urn. The remaining marbles
inside the urn are again mixed and one marble is picked at random. This kind of
selection process is often referred to as sampling without replacement. Now,
what is the probability that (i) both the first and second drawn marbles are
green, (ii) the second drawn marble is green? Let us define the two events
A: The randomly selected first marble is green
B: The randomly selected second marble is green

Obviously, P(A) = 8/20 = .4 and P (B | A) = 7/19. Observe that the experimen-
tal setup itself dictates the value of P(B | A). A result such as (1.4.3) is not very
helpful in the present situation. In order to answer question (i), we proceed by
using (1.4.4) to evaluate P(A ∩ B) = P(A) P(B | A) = 8/20 7/19 = 14/95. Obvi-
c
ously, {A, A } forms a partition of the sample space. Now, in order to answer
question (ii), using the Theorem 1.3.1, part (vi), we write

But, as before we have P(A ∩ B) = P(A P(B | A ) = 12/20 8/19 = 24/95. Thus,
c
c
c
from (1.4.5), we have P(B) = 14/95 + 24/95 = 38/95 = 2/5 = .4. Here, note that
P(B | A) ≠ P(B) and so by the Definition 1.4.2, the two events A, B are depen-
dent. One may guess this fact easily from the layout of the experiment itself.
The reader should check that P(A) would be equal to P(B) whatever, be the
configuration of the urn. Refer to the Exercise 1.4.11. !

In the Example 1.4.4, P(B | A) was known to us, and so we
could use (1.4.4) to find the joint probability P(A ∩ B).