Study Notes on the Relationship between Methamphetamine Use and Paranoia

The study addresses the relationship between methamphetamine use and paranoia in individuals.

To calculate the percentage of individuals who are both Methamphetamine-users and Paranoid individuals, we utilize the data provided.
The formula to find the probability of two events occurring is: $P(A ext{ and } B) = P(B|A) \times P(A)$ Where:
- A = Being a Methamphetamine-user
- B = Being a Paranoid individual
Given:
- P(B|A) (probability of being Paranoid given that one is a Methamphetamine-user) = 96% = 0.96
- P(A) (probability of being a Methamphetamine-user) = 3% = 0.03
Therefore,
$P(A ext{ and } B) = 0.96 imes 0.03 = 0.0288$
This means 2.88% of the general population are both Methamphetamine-users and Paranoid individuals.

To find the probability that a Paranoid individual is a Methamphetamine-user, we apply Bayes' theorem:
$P(A|B) = \frac{P(B|A)\times P(A)}{P(B)}$
To solve this, the probabilities are as follows:
- P(B) (probability of being Paranoid) = 7% = 0.07
Substituting the known values into Bayes' theorem:
$P(A|B) = \frac{0.96 \times 0.03}{0.07} = \frac{0.0288}{0.07} \approx 0.4114$
Thus, approximately 41.14% of Paranoid individuals are Methamphetamine-users.

To determine if the events are independent, we check the following condition:
- Events A and B are independent if and only if
  $P(A ext{ and } B) = P(A) \times P(B)$
Given:
From previous calculations:
- P(A) = 3% = 0.03
- P(B) = 7% = 0.07
- P(A ext{ and } B) = 2.88% = 0.0288
Now, calculate $P(A) \times P(B)$ :
$0.03 \times 0.07 = 0.0021$
Comparison:
- P(A ext{ and } B) = 0.0288
- P(A) \times P(B) = 0.0021
Since 0.0288 is not equal to 0.0021, it follows that the two events are not independent.