Bayes' Theorem
and that the fair coin tossed is a Bernoulli trial. Thus we have
and we have . Hence we get
When is changed to 1/1000, we will have
and when
is changed to
1/50 we will have
.
Independence
solving for N yields N=24. The claim that the race and sex of the people on their staff are independent is not equivalent to saying that they hire people independent of their race and sex, since the demographics of their staff may not reflect the demographics of their applicant pool.
Hence we have . Therefore
and C are independent.
Since ,
for any
i,
. Thus each
is either a singleton
set or a doubleton set. Clearly,
for
, since if
, then we will have,
, and in this case
and
will not be independent of each other, contradicting
our
assumption.
Now, for any pair and
there are only two
possibilities: Either
, or
. If
, then we will
have,
, and this will make
and
not independent of
each other, again contradicting
our assumption.
If , since
and
are both true,
both
and
should be doubleton
sets such that
for some
,
,
and
for distinct
and k. In this case, the following will hold true:
and, since and
are independent,
Thus we get and hence
, which is
impossible since
is true. Hence, it is
impossible for
and
to be pairwise independent.
Let be the events in Problem 8 above. We know that
A and B are independent, and
P(C) > 0 and we have
.
But,
. Thus we prove that
is not necessarily equal to
.