The period of is
.
Recall that and
both have periods of
. This means
Replacing with
, we have
In other words, if we change by some multiple of
, we end up with the same output. So
has period
.
Similarly, has a period of
,
We want to find the period of
, such that
On the left side, we have
and
So, in terms of its period, we have
and we need to find the smallest positive such that
which points to , since
Solution :
It is given that :
P (positive | Has disease) = 0.7
P (positive | No disease) = 0.08
P (has disease) = 0.18
P (No disease) = 1 - 0.18
= 0.82
Now if test administered to the individual is positive, the probability that the person actually have the disease is
P (Has disease | positive) ......(1)
The P(positive) is,
= P(positive | has disease) x P(Has disease) + P(positive | no disease) x P(No disease)
= 0.7 (0.19) + 0.04 (0.81)
= 0.1654
Now substituting the values in the equation (1), we get
P (Has disease | positive)
= 0.8041